## Question

Determine whether the series \(\sum\limits_{n = 1}^\infty {\frac{{{n^{10}}}}{{{{10}^n}}}} \) is convergent or divergent.

**Answer/Explanation**

## Markscheme

Consider

\(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{{{(n + 1)}^{10}}}}{{{{10}^{n + 1}}}} \times \frac{{{{10}^n}}}{{{n^{10}}}}\) **M1A1**

\( = \frac{1}{{10}}{\left( {1 + \frac{1}{n}} \right)^{10}}\) **A1**

\( \to \frac{1}{{10}}{\text{ as }}n \to \infty \) *A1*

\(\frac{1}{{10}} < 1\) *R1*

So by the Ratio Test the series is convergent. *R1*

*[6 marks]*

## Examiners report

Most candidates used the Ratio Test successfully to establish convergence. Candidates who attempted to use Cauchy’s (Root) Test were often less successful although this was a valid method.

## Question

Find the radius of convergence of the series \(\sum\limits_{n = 0}^\infty {\frac{{{{( – 1)}^n}{x^n}}}{{(n + 1){3^n}}}} \).

Determine whether the series \(\sum\limits_{n = 0}^\infty {\left( {\sqrt[3]{{{n^3} + 1}} – n} \right)} \) is convergent or divergent.

**Answer/Explanation**

## Markscheme

The ratio test gives

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( – 1)}^{n + 1}}{x^{n + 1}}(n + 1){3^n}}}{{(n + 2){3^{n + 1}}{{( – 1)}^n}{x^n}}}} \right|\) *M1A1*

\( = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1)x}}{{3(n + 2)}}} \right|\) *A1*

\( = \frac{{\left| x \right|}}{3}\) *A1*

So the series converges for \( \frac{{\left| x \right|}}{3} < 1,\) *A1*

the radius of convergence is 3 *A1*

**Note: **Do not penalise lack of modulus signs.

*[6 marks]*

\({u_n} = \sqrt[3]{{{n^3} + 1}} – n\)

\( = n\left( {\sqrt[3]{{1 + \frac{1}{{{n^3}}} – 1}}} \right)\) **M1A1**

\( = n\left( {1 + \frac{1}{{3{n^3}}} – \frac{1}{{9{n^6}}} + \frac{5}{{8{\text{l}}{n^9}}} – … – 1} \right)\) **A1**

using \({v_n} = \frac{1}{{{n^2}}}\) as the auxilliary series, *M1*

since \(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{{v_n}}} = \frac{1}{3}{\text{ and }}\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + …\) converges *M1A1*

then \(\sum {{u_n}} \) converges *A1*

**Note: **Award ** M1A1A1M0M0A0A0 **to candidates attempting to use the integral test.

*[7 marks]*

## Examiners report

Some corners were cut in applying the ratio test and some candidates tried to use the comparison test. With careful algebra finding the radius of convergence was not too difficult. Often the interval of convergence was given instead of the radius.

Part (b) was done only by the best candidates. A little algebraic manipulation together with an auxiliary series soon gave the answer.

Some corners were cut in applying the ratio test and some candidates tried to use the comparison test. With careful algebra finding the radius of convergence was not too difficult. Often the interval of convergence was given instead of the radius.

Part (b) was done only by the best candidates. A little algebraic manipulation together with an auxiliary series soon gave the answer.

## Question

(i) Show that \(\int_1^\infty {\frac{1}{{x(x + p)}}{\text{d}}x,{\text{ }}p \ne 0} \) is convergent if *p *> −1 and find its value in terms of *p*.

(ii) Hence show that the following series is convergent.

\[\frac{1}{{1 \times 0.5}} + \frac{1}{{2 \times 1.5}} + \frac{1}{{3 \times 2.5}} + …\]

Determine, for each of the following series, whether it is convergent or divergent.

(i) \(\sum\limits_{n = 1}^\infty {\sin \left( {\frac{1}{{n(n + 3)}}} \right)} \)

(ii) \(\sqrt {\frac{1}{2}} + \sqrt {\frac{1}{6}} + \sqrt {\frac{1}{{12}}} + \sqrt {\frac{1}{{20}}} + …\)

**Answer/Explanation**

## Markscheme

(i) the integrand is non-singular on the domain if *p* > –1 with the latter assumed, consider

\(\int_1^R {\frac{1}{{x(x + p)}}} {\text{d}}x = \frac{1}{p}\int_1^R {\frac{1}{x} – \frac{1}{{x + p}}{\text{d}}x} \) *M1A1*

\( = \frac{1}{p}\left[ {\ln \left( {\frac{x}{{x + p}}} \right)} \right]_1^R,{\text{ }}p \ne 0\) *A1*

this evaluates to

\(\frac{1}{p}\left( {\ln \frac{R}{{R + p}} – \ln \frac{1}{{1 + p}}} \right),{\text{ }}p \ne 0\) *M1*

\( \to \frac{1}{p}\ln (1 + p)\) *A1*

because \(\frac{R}{{R + p}} \to 1{\text{ as }}R \to \infty \) *R1*

hence the integral is convergent *AG*

* *

(ii) the given series is \(\sum\limits_{n = 1}^\infty {f(n),{\text{ }}f(n) = \frac{1}{{n(n – 0.5)}}} \) *M1*

the integral test and *p* = –0.5 in (i) establishes the convergence of the series *R1*

*[8 marks]*

(i) as we have a series of positive terms we can apply the comparison test, limit form

comparing with \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) *M1*

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {\frac{1}{{n(n + 3)}}} \right)}}{{\frac{1}{{{n^2}}}}} = 1\) *M1A1*

as \(\sin \theta \approx \theta {\text{ for small }}\theta \) *R1*

and \(\frac{{{n^2}}}{{n(n + 3)}} \to 1\) *R1*

(so as the limit (of 1) is finite and non-zero, both series exhibit the same behavior)

\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) converges, so this series converges *R1*

* *

(ii) the general term is

\(\sqrt {\frac{1}{{n(n + 1)}}} \) *A1*

\(\sqrt {\frac{1}{{n(n + 1)}}} > \sqrt {\frac{1}{{(n + 1)(n + 1)}}} \) *M1*

\(\sqrt {\frac{1}{{(n + 1)(n + 1)}}} = \frac{1}{{n + 1}}\) *A1*

the harmonic series diverges *R1*

so by the comparison test so does the given series *R1*

*[11 marks]*

## Examiners report

Part(a)(i) caused problems for some candidates who failed to realize that the integral can only be tackled by the use of partial fractions. Even then, the improper integral only exists as a limit – too many candidates ignored or skated over this important point. Candidates must realize that in this type of question, rigour is important, and full marks will only be awarded for a full and clearly explained argument. This applies as well to part(b), where it was also noted that some candidates were confusing the convergence of the terms of a series to zero with convergence of the series itself.

Part(a)(i) caused problems for some candidates who failed to realize that the integral can only be tackled by the use of partial fractions. Even then, the improper integral only exists as a limit – too many candidates ignored or skated over this important point. Candidates must realize that in this type of question, rigour is important, and full marks will only be awarded for a full and clearly explained argument. This applies as well to part(b), where it was also noted that some candidates were confusing the convergence of the terms of a series to zero with convergence of the series itself.

## Question

Determine whether the series \(\sum\limits_{n = 1}^\infty {\sin \frac{1}{n}} \) is convergent or divergent.

Show that the series \(\sum\limits_{n = 2}^\infty {\frac{1}{{n{{(\ln n)}^2}}}} \) is convergent.

**Answer/Explanation**

## Markscheme

comparing with the series \(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \) *A1*

using the limit comparison test *(M1)*

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \frac{1}{n}}}{{\frac{1}{n}}}\left( { = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}} \right) = 1\) *M1A1*

since \(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \) diverges, \(\sum\limits_{n = 1}^\infty {\sin \frac{1}{n}} \) diverges *A1*

*[5 marks]*

using integral test *(M1)*

let \(u = \ln x\) *(M1)*

\( \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}\)

\(\int {\frac{1}{{x{{(\ln x)}^2}}}{\text{d}}x = \int {\frac{1}{{{u^2}}}{\text{d}}u = – \frac{1}{u}\left( { = – \frac{1}{{\ln x}}} \right)} } \) *A1*

\( \Rightarrow \int_2^\infty {\frac{1}{{x{{(\ln x)}^2}}}{\text{d}}x = \mathop {\lim }\limits_{a \to \infty } } \left[ { – \frac{1}{{\ln x}}} \right]_2^a\)

\( = \mathop {\lim }\limits_{a \to \infty } \left( { – \frac{1}{{\ln a}} + \frac{1}{{\ln 2}}} \right)\) *(M1)(A1)*

as \(a \to \infty ,{\text{ }} – \frac{1}{{\ln a}} \to 0\) *A1*

\( \Rightarrow \int_2^\infty {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} {\text{d}}x = \frac{1}{{\ln 2}}\)

hence the series is convergent *AG*

*[7 marks]*

## Examiners report

This question was found to be the hardest on the paper, with only the best candidates gaining full marks on it. Part (a) was very poorly done with a significant number of candidates unable to start the question. More students recognised part (b) as an integral test, but often could not progress beyond this. In many cases, students appeared to be guessing at what might constitute a valid test.

This question was found to be the hardest on the paper, with only the best candidates gaining full marks on it. Part (a) was very poorly done with a significant number of candidates unable to start the question. More students recognised part (b) as an integral test, but often could not progress beyond this. In many cases, students appeared to be guessing at what might constitute a valid test.

## Question

Find the radius of convergence of the infinite series

\[\frac{1}{2}x + \frac{{1 \times 3}}{{2 \times 5}}{x^2} + \frac{{1 \times 3 \times 5}}{{2 \times 5 \times 8}}{x^3} + \frac{{1 \times 3 \times 5 \times 7}}{{2 \times 5 \times 8 \times 11}}{x^4} + \ldots {\text{ .}}\]

Determine whether the series \(\sum\limits_{n = 1}^\infty {\sin \left( {\frac{1}{n} + n\pi } \right)} \) is convergent or divergent.

**Answer/Explanation**

## Markscheme

the *n*th term is

\({u_n} = \frac{{1 \times 3 \times 5 \ldots (2n – 1)}}{{2 \times 5 \times 8 \ldots (3n – 1)}}{x^n}\) *M1A1*

(using the ratio test to test for absolute convergence)

\(\frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{(2n + 1)}}{{(3n + 2)}}\left| x \right|\) *M1A1*

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{2\left| x \right|}}{3}\) *A1*

let *R* denote the radius of convergence

then \(\frac{{2R}}{3} = 1\) so \(r = \frac{3}{2}\) *M1A1*

**Note:** Do not penalise the absence of absolute value signs.

*[7 marks]*

using the compound angle formula or a graphical method the series can be written in the form *(M1)*

\(\sum\limits_{n = 1}^\infty {{u_n}} \) where \({u_n} = {( – 1)^n}\sin \left( {\frac{1}{n}} \right)\) *A2*

since \(\frac{1}{n} < \frac{\pi }{2}\) *i.e.* an angle in the first quadrant, *R1*

it is an alternating series *R1*

\({u_n} \to 0{\text{ as }}n \to \infty \) *R1*

and \(\left| {{u_{n + 1}}} \right| < \left| {{u_n}} \right|\) *R1*

it follows that the series is convergent *R1*

*[8 marks]*

## Examiners report

Solutions to this question were generally disappointing. In (a), many candidates were unable even to find an expression for the *n*th term so that they could not apply the ratio test.

Solutions to this question were generally disappointing. In (b), few candidates were able to rewrite the *n*th term in the form \(\sum {{{( – 1)}^n}\sin \left( {\frac{1}{n}} \right)} \) so that most candidates failed to realise that the series was alternating.

## Question

Consider the power series \(\sum\limits_{k = 1}^\infty {k{{\left( {\frac{x}{2}} \right)}^k}} \).

(i) Find the radius of convergence.

(ii) Find the interval of convergence.

Consider the infinite series \(\sum\limits_{k = 1}^\infty {{{( – 1)}^{k + 1}} \times \frac{k}{{2{k^2} + 1}}} \).

(i) Show that the series is convergent.

(ii) Show that the sum to infinity of the series is less than 0.25.

**Answer/Explanation**

## Markscheme

(i) consider \(\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{\left| {\frac{{(n – 1){x^{n + 1}}}}{{{2^{n + 1}}}}} \right|}}{{\left| {\frac{{n{x^n}}}{{{2^n}}}} \right|}}\) *M1*

\( = \frac{{(n + 1)\left| x \right|}}{{2n}}\) *A1*

\( \to \frac{{\left| x \right|}}{2}{\text{ as }}n \to \infty \) *A1*

the radius of convergence satisfies

\(\frac{R}{2} = 1\), *i.e. R* = 2 *A1*

* *

(ii) the series converges for \( – 2 < x < 2\), we need to consider \(x = \pm 2\) *(R1)*

when *x* = 2, the series is \(1 + 2 + 3 + \ldots \) *A1*

this is divergent for any one of several reasons *e.g.* finding an expression for or a comparison test with the harmonic series or noting that \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\) etc. *R1*

when *x* = – 2, the series is \( – 1 + 2 – 3 + 4 \ldots \) *A1*

this is divergent for any one of several reasons

*e.g.* partial sums are

\( – 1,{\text{ }}1,{\text{ }} – 2,{\text{ }}2,{\text{ }} – 3,{\text{ }}3 \ldots \) or noting that \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\) *etc.* *R1*

the interval of convergence is \( – 2 < x < 2\) *A1*

*[10 marks]*

(i) this alternating series is convergent because the moduli of successive terms are monotonic decreasing *R1*

and the \({n^{{\text{th}}}}\) term tends to zero as \(n \to \infty \) *R1*

* *

(ii) consider the partial sums

0.333, 0.111, 0.269, 0.148, 0.246 *M1A1*

since the sum to infinity lies between any pair of successive partial sums, it follows that the sum to infinity lies between 0.148 and 0.246 so that it is less than 0.25 *R1*

**Note:** Accept a solution which looks only at 0.333, 0.269, 0.246 and states that these are successive upper bounds.

*[5 marks]*

## Examiners report

Most candidates found the radius of convergence correctly but examining the situation when \(x = \pm 2\) often ended in loss of marks through inadequate explanations. In (b)(i) many candidates were able to justify the convergence of the given series. In (b)(ii), however, many candidates seemed unaware of the fact the sum to infinity lies between any pair of successive partial sums.

Most candidates found the radius of convergence correctly but examining the situation when \(x = \pm 2\) often ended in loss of marks through inadequate explanations. In (b)(i) many candidates were able to justify the convergence of the given series. In (b)(ii), however, many candidates seemed unaware of the fact the sum to infinity lies between any pair of successive partial sums.

## Question

Determine whether or not the following series converge.

(a) \(\sum\limits_{n = 0}^\infty {\left( {\sin \frac{{n\pi }}{2} – \sin \frac{{(n + 1)\pi }}{2}} \right)} \)

(b) \(\sum\limits_{n = 1}^\infty {\frac{{{{\text{e}}^n} – 1}}{{{\pi ^n}}}} \)

(c) \(\sum\limits_{n = 2}^\infty {\frac{{\sqrt {n + 1} }}{{n(n – 1)}}} \)

**Answer/Explanation**

## Markscheme

(a) \(\sum\limits_{n = 0}^\infty {\left( {\sin \frac{{n\pi }}{2} – \sin \frac{{(n + 1)\pi }}{2}} \right)} \)

\( = \left( {\sin 0 – \sin \frac{\pi }{2}} \right) + \left( {\sin \frac{\pi }{2} – \sin \pi } \right) + \left( {\sin \pi – \sin \frac{{3\pi }}{2}} \right) + \left( {\sin \frac{{3\pi }}{2} – \sin 2\pi } \right) + \ldots \) *(M1)*

the \({n^{{\text{th}}}}\) term is ±1 for all *n*, *i.e.* the \({n^{{\text{th}}}}\) term does not tend to 0 *A1*

hence the series does not converge *A1*

*[3 marks]*

* *

(b) **EITHER**

using the ratio test *(M1)*

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\text{e}}^{n + 1}}}}{{{\pi ^{n + 1}}}}} \right)\left( {\frac{{{\pi ^n}}}{{{{\text{e}}^n} – 1}}} \right)\) *M1A1*

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\text{e}}^{n + 1}} – 1}}{{{{\text{e}}^n} – 1}}} \right)\left( {\frac{{{\pi ^n}}}{{{\pi ^{n + 1}}}}} \right) = \frac{{\text{e}}}{\pi }\,\,\,\,\,( \approx 0.865)\) *M1A1*

\(\frac{{\text{e}}}{\pi } < 1\), hence the series converges *R1A1*

**OR**

\(\sum\limits_{n = 1}^\infty {\frac{{{{\text{e}}^n} – 1}}{{{\pi ^n}}} = \sum\limits_{n = 1}^\infty {{{\left( {\frac{{\text{e}}}{\pi }} \right)}^n} – {{\left( {\frac{1}{\pi }} \right)}^n} = \sum\limits_{n = 1}^\infty {{{\left( {\frac{{\text{e}}}{\pi }} \right)}^n} – \sum\limits_{n = 1}^\infty {{{\left( {\frac{{\text{1}}}{\pi }} \right)}^n}} } } } \) *M1A1*

the series is the difference of two geometric series, with \(r = \frac{{\text{e}}}{\pi }\,\,\,\,\,( \approx 0.865)\) *M1A1*

and \(\frac{1}{\pi }\,\,\,\,\,( \approx 0.318)\) *A1*

for both \(\left| r \right| < 1\), hence the series converges *R1A1*

**OR**

\(\forall n,{\text{ }}0 < \frac{{{{\text{e}}^n} – 1}}{{{\pi ^n}}} < \frac{{{{\text{e}}^n}}}{{{\pi ^n}}}\) *(M1)A1A1*

the series \(\frac{{{{\text{e}}^n}}}{{{\pi ^n}}}\) converges since it is a geometric series such that \(\left| r \right| < 1\) *A1R1*

therefore, by the comparison test, \(\frac{{{{\text{e}}^n} – 1}}{{{\pi ^n}}}\) converges *R1A1*

*[7 marks]*

* *

(c) by limit comparison test with \(\frac{{\sqrt n }}{{{n^2}}}\), *(M1)*

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\frac{{\sqrt {n + 1} }}{{n(n – 1)}}}}{{\frac{{\sqrt n }}{{{n^2}}}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\sqrt {n + 1} }}{{n(n – 1)}} \times \frac{{{n^2}}}{{\sqrt n }}} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n – 1}}\sqrt {\frac{{n + 1}}{n}} = 1\) *M1A1*

hence both series converge or both diverge *R1*

by the *p*-test \(\sum\limits_{n = 1}^\infty {\frac{{\sqrt n }}{{{n^2}}} = {n^{\frac{{ – 3}}{2}}}} \) converges, hence both converge *R1A1*

*[6 marks]*

*Total [16 marks]*

## Examiners report

This was the least successfully answered question on the paper. Candidates often did not know which convergence test to use; hence very few full successful solutions were seen. The communication of the method used was often quite poor.

a) Many candidates failed to see that this is a telescoping series. If this was recognized then the question was fairly straightforward. Often candidates unsuccessfully attempted to apply the standard convergence tests.

b) Many candidates used the ratio test, but some had difficulty in simplifying the expression. Others recognized that the series is the difference of two geometric series, and although the algebraic work was done correctly, some failed to communicate the conclusion that since the absolute value of the ratios are less than 1, hence the series converges. Some candidates successfully used the comparison test.

c) Although the limit comparison test was attempted by most candidates, it often failed through an inappropriate selection of a series.

## Question

Consider the infinite series

\[\frac{1}{{2\ln 2}} – \frac{1}{{3\ln 3}} + \frac{1}{{4\ln 4}} – \frac{1}{{5\ln 5}} + \ldots {\text{ .}}\]

(a) Show that the series converges.

(b) Determine if the series converges absolutely or conditionally.

**Answer/Explanation**

## Markscheme

(a) applying the alternating series test as \(\forall n \geqslant 2,\frac{1}{{n\ln n}} \in {\mathbb{R}^ + }\) *M1*

\(\forall n,\frac{1}{{(n + 1)\ln (n + 1)}} \leqslant \frac{1}{{n\ln n}}\) *A1*

\(\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n\ln n}} = 0\) *A1*

hence, by the alternating series test, the series converges *R1*

*[4 marks]*

* *

(b) as \(\frac{1}{{x\ln x}}\) is a continuous decreasing function, apply the integral test to determine if it converges absolutely *(M1)*

\(\int_2^\infty {\frac{1}{{x\ln x}}{\text{d}}x = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{1}{{x\ln x}}{\text{d}}x} } \) *M1A1*

let \(u = \ln x\) then \({\text{d}}u = \frac{1}{x}{\text{d}}x\) *(M1)A1*

\(\int {\frac{1}{u}{\text{d}}u = \ln u} \) *(A1)*

hence, \(\mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{1}{{x\ln x}}{\text{d}}x = \mathop {\lim }\limits_{b \to \infty } \left[ {\ln (\ln x)} \right]_2^b} \) which does not exist *M1A1A1*

hence, the series does not converge absolutely *(A1)*

the series converges conditionally *A1*

*[11 marks]*

*Total [15 marks]*

## Examiners report

Part (a) was answered well by many candidates who attempted this question. In part (b), those who applied the integral test were mainly successful, but too many failed to supply the justification for its use, and proper conclusions.

## Question

The exponential series is given by \({{\text{e}}^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \) .

Find the set of values of *x* for which the series is convergent.

(i) Show, by comparison with an appropriate geometric series, that

\[{{\text{e}}^x} – 1 < \frac{{2x}}{{2 – x}},{\text{ for }}0 < x < 2{\text{.}}\]

(ii) Hence show that \({\text{e}} < {\left( {\frac{{2n + 1}}{{2n – 1}}} \right)^n}\), for \(n \in {\mathbb{Z}^ + }\).

(i) Write down the first three terms of the Maclaurin series for \(1 – {{\text{e}}^{ – x}}\) and explain why you are able to state that

\[1 – {{\text{e}}^{ – x}} > x – \frac{{{x^2}}}{2},{\text{ for }}0 < x < 2.\]

(ii) Deduce that \({\text{e}} > {\left( {\frac{{2{n^2}}}{{2{n^2} – 2n + 1}}} \right)^n}\), for \(n \in {\mathbb{Z}^ + }\).

Letting *n* = 1000, use the results in parts (b) and (c) to calculate the value of e correct to as many decimal places as possible.

**Answer/Explanation**

## Markscheme

using a ratio test,

\(\left| {\frac{{{T_{n + 1}}}}{{{T_n}}}} \right| = \left| {\frac{{{x^{n + 1}}}}{{(n + 1)!}}} \right| \times \left| {\frac{{n!}}{{{x^n}}}} \right| = \frac{{\left| x \right|}}{{n + 1}}\) *M1A1*

**Note:** Condone omission of modulus signs.

\( \to 0{\text{ as }}n \to \infty \) for all values of *x* *R1*

the series is therefore convergent for \(x \in \mathbb{R}\) *A1*

*[4 marks]*

(i) \({{\text{e}}^x} – 1 = x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{{2 \times 3}} + …\) *M1*

\( < x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{{2 \times 2}} + …\,\,\,\,\,({\text{for }}x > 0)\) *A1*

\( = \frac{x}{{1 – \frac{x}{2}}}\,\,\,\,\,({\text{for }}x < 2)\) *A1*

\( = \frac{{2x}}{{2 – x}}\,\,\,\,\,({\text{for }}0 < x < 2)\) *AG*

* *

(ii) \({{\text{e}}^x} < 1 + \frac{{2x}}{{2 – x}} = \frac{{2 + x}}{{2 – x}}\) *A1*

\({\text{e}} < {\left( {\frac{{2 + x}}{{2 – x}}} \right)^{\frac{1}{x}}}\) *A1*

replacing *x* by \(\frac{1}{n}\) (and noting that the result is true for \(n > \frac{1}{2}\) and therefore \({\mathbb{Z}^ + }\) ) *M1*

\({\text{e}} < {\left( {\frac{{2n + 1}}{{2n – 1}}} \right)^n}\) *AG*

*[6 marks]*

(i) \(1 – {{\text{e}}^{ – x}} = x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + …\) *A1*

for \(0 < x < 2\), the series is alternating with decreasing terms so that the sum is greater than the sum of an even number of terms *R1*

therefore

\(1 – {{\text{e}}^{ – x}} > x – \frac{{{x^2}}}{2}\) *AG*

* *

(ii) \({{\text{e}}^{ – x}} < 1 – x + \frac{{{x^2}}}{2}\)

\({{\text{e}}^x} > \frac{1}{{\left( {1 – x + \frac{{{x^2}}}{2}} \right)}}\) *M1*

\({\text{e}} > {\left( {\frac{2}{{2 – 2x + {x^2}}}} \right)^{\frac{1}{x}}}\) *A1*

replacing *x* by \(\frac{1}{n}\) (and noting that the result is true for \(n > \frac{1}{2}\) and therefore \({\mathbb{Z}^ + }\) )

\({\text{e}} > {\left( {\frac{{2{n^2}}}{{2{n^2} – 2n + 1}}} \right)^n}\) *AG*

* *

*[4 marks]*

from (b) and (c), \({\text{e}} < 2.718282…\) and \({\text{e}} > 2.718281…\) *A1*

we conclude that e = 2.71828 correct to 5 decimal places *A1*

*[2 marks]*

## Examiners report

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

## Question

Show that \(n! \geqslant {2^{n – 1}}\), for \(n \geqslant 1\).

Hence use the comparison test to determine whether the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{n!}}} \) converges or diverges.

**Answer/Explanation**

## Markscheme

for \(n \geqslant 1,{\text{ }}n! = n(n – 1)(n – 2) \ldots 3 \times 2 \times 1 \geqslant 2 \times 2 \times 2 \ldots 2 \times 2 \times 1 = {2^{n – 1}}\) *M1A1*

\( \Rightarrow n! \geqslant {2^{n – 1}}{\text{ for }}n \geqslant 1\) *AG*

*[2 marks]*

\(n! \geqslant {2^{n – 1}} \Rightarrow \frac{1}{{n!}} \leqslant \frac{1}{{{2^{n – 1}}}}{\text{ for }}n \geqslant 1\) *A1*

\(\sum\limits_{n = 1}^\infty {\frac{1}{{{2^{n – 1}}}}} \) is a positive converging geometric series *R1*

hence \(\sum\limits_{n = 1}^\infty {\frac{1}{{n!}}} \) converges by the comparison test *R1*

*[3 marks]*

## Examiners report

Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’ being that the result is true for *n* = 1, 2, 3 and therefore true for all *n*. Some candidates attempted to use induction which is a valid method but no completely correct solution using this method was seen. Candidates found part (b) more accessible and many correct solutions were seen. The most common problem was candidates using an incorrect comparison test, failing to realise that what was required was a comparison between \(\sum {\frac{1}{{n!}}} \) and \(\sum {\frac{1}{{{2^{n – 1}}}}} \).

Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’ being that the result is true for *n* = 1, 2, 3 and therefore true for all *n*. Some candidates attempted to use induction which is a valid method but no completely correct solution using this method was seen. Candidates found part (b) more accessible and many correct solutions were seen. The most common problem was candidates using an incorrect comparison test, failing to realise that what was required was a comparison between \(\sum {\frac{1}{{n!}}} \) and \(\sum {\frac{1}{{{2^{n – 1}}}}} \).

## Question

Using the integral test, show that \(\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} \) is convergent.

(i) Show, by means of a diagram, that \(\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} < \frac{1}{{4 \times {1^2} + 1}} + \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x} \).

(ii) Hence find an upper bound for \(\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} \)

**Answer/Explanation**

## Markscheme

\(\int {\frac{1}{{4{x^2} + 1}}{\text{d}}x = \frac{1}{2}\arctan 2x + k} \) *(M1)(A1)*

**Note:** Do not penalize the absence of “+*k*”.

\(\int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x = \frac{1}{2}\mathop {\lim }\limits_{a \to \infty } } [\arctan 2x]_1^a\) *(M1)*

**Note:** Accept \(\frac{1}{2}[\arctan 2x]_1^\infty \).

\( = \frac{1}{2}\left( {\frac{\pi }{2} – \arctan 2} \right)\,\,\,\,\,( = 0.232)\) *A1*

hence the series converges *AG*

*[4 marks]*

(i)

* A2*

The shaded rectangles lie within the area below the graph so that \(\sum\limits_{n = 2}^\infty {\frac{1}{{4{n^2} + 1}}} < \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x} \). Adding the first term in the series, \(\frac{1}{{4 \times {1^2} + 1}}\), gives \(\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} < \frac{1}{{4 \times {1^2} + 1}} + \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x} \) *R1AG*

* *

(ii) upper bound \( = \frac{1}{5} + \frac{1}{2}\left( {\frac{\pi }{2} – \arctan 2} \right)\,\,\,\,\,( = 0.432)\) *A1*

*[4 marks]*

## Examiners report

This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate \({\frac{1}{{4{x^2} + 1}}}\) successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.

This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate \({\frac{1}{{4{x^2} + 1}}}\) successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.

## Question

The sequence \(\{ {u_n}\} \) is defined by \({u_n} = \frac{{3n + 2}}{{2n – 1}}\), for \(n \in {\mathbb{Z}^ + }\).

Show that the sequence converges to a limit *L *, the value of which should be stated.

Find the least value of the integer *N *such that \(\left| {{u_n} – L} \right| < \varepsilon \) , for all *n *> *N *where

(i) \(\varepsilon = 0.1\);

(ii) \(\varepsilon = 0.00001\).

For each of the sequences \(\left\{ {\frac{{{u_n}}}{n}} \right\},{\text{ }}\left\{ {\frac{1}{{2{u_n} – 2}}} \right\}\) and \(\left\{ {{{( – 1)}^n}{u_n}} \right\}\) , determine whether or not it converges.

Prove that the series \(\sum\limits_{n = 1}^\infty {({u_n} – L)} \) diverges.

**Answer/Explanation**

## Markscheme

\({u_n} = \frac{{3 + \frac{2}{n}}}{{2 – \frac{1}{n}}}\) or \(\frac{3}{2} + \frac{A}{{2n – 1}}\) *M1*

using \(\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0\) *(M1)*

obtain \(\mathop {\lim }\limits_{n \to \infty } {u_n} = \frac{3}{2} = L\) *A1 N1*

*[3 marks]*

\({u_n} – L = \frac{7}{{2(2n – 1)}}\) ** (A1)**

\(\left| {{u_n} – L} \right| < \varepsilon \Rightarrow n > \frac{1}{2}\left( {1 + \frac{7}{{2\varepsilon }}} \right)\) ** (M1)**

* *

(i) \(\varepsilon = 0.1 \Rightarrow N = 18\) *A1*

* *

(ii) \(\varepsilon = 0.00001 \Rightarrow N = 175000\) *A1*

*[4 marks]*

\({u_n} \to L\) and \(\frac{1}{n} \to 0\) *M1*

\( \Rightarrow \frac{{{u_n}}}{n} \to (L \times 0) = 0\) , hence converges *A1*

\(2{u_n} – 2 \to 2L – 2 = 1 \Rightarrow \frac{1}{{2{u_n} – 2}} \to 1\) , hence converges *M1A1*** **

**Note: **To award ** A1 **the value of the limit and a statement of convergence must be clearly seen for each sequence.

\({( – 1)^n}{u_n}\) does not converge *A1*

The sequence alternates (or equivalent wording) between values close to \( \pm L\) *R1*

*[6 marks]*

\({u_n} – L > \frac{7}{{4n}}\) (re: harmonic sequence) *M1*

\( \Rightarrow \sum\limits_{n = 1}^\infty {({u_n} – L)} \) diverges by the comparison theorem *R1*** **

**Note: **Accept alternative methods.

*[2 marks]*

## Examiners report

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

## Question

Find the set of values of *k *for which the improper integral \(\int_2^\infty {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \) converges.

Show that the series \(\sum\limits_{r = 2}^\infty {\frac{{{{( – 1)}^r}}}{{r\ln r}}} \) is convergent but not absolutely convergent.

**Answer/Explanation**

## Markscheme

consider the limit as \(R \to \infty \) of the (proper) integral

\(\int_2^R {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \) ** (M1)**

substitute \(u = \ln x,{\text{ d}}u = \frac{1}{x}{\text{d}}x\) (*M1)*

obtain \(\int_{\ln 2}^{\ln R} {\frac{1}{{{u^k}}}{\text{d}}u = \left[ { – \frac{1}{{k – 1}}\frac{1}{{{u^{k – 1}}}}} \right]_{\ln 2}^{\ln R}} \) *A1*** **

**Note: **Ignore incorrect limits or omission of limits at this stage.

or \([\ln u]_{\ln 2}^{\ln R}\) if *k* = 1 *A1*** **

**Note: **Ignore incorrect limits or omission of limits at this stage.

because \(\ln R{\text{ }}({\text{and }}\ln \ln R) \to \infty {\text{ as }}R \to \infty \) *(M1)*

converges in the limit if *k* > 1 *A1*

*[6 marks]*

C: \({\text{terms}} \to 0{\text{ as }}r \to \infty \) *A1*

\(\left| {{u_{r + 1}}} \right| < \left| {{u_r}} \right|\) for all *r **A1*

convergence by alternating series test *R1*

AC: \({(x\ln x)^{ – 1}}\) is positive and decreasing on \([2,\,\infty )\) *A1*

not absolutely convergent by integral test using part (a) for *k* = 1 *R1*

*[5 marks]*

## Examiners report

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when *k* = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when *k* = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

## Question

Prove that \(\mathop {\lim }\limits_{H \to \infty } \int_a^H {\frac{1}{{{x^2}}}{\text{d}}x} \) exists and find its value in terms of \(a{\text{ (where }}a \in {\mathbb{R}^ + })\).

Use the integral test to prove that \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) converges.

Let \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = L\) .

The diagram below shows the graph of \(y = \frac{1}{{{x^2}}}\).

(i) Shade suitable regions on a copy of the diagram above and show that

\(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_{k + 1}^\infty {\frac{1}{{{x^2}}}} {\text{d}}x < L\) .

(ii) Similarly shade suitable regions on another copy of the diagram above and

show that \(L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_k^\infty {\frac{1}{{{x^2}}}} {\text{d}}x\) .

Hence show that \(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{{k + 1}} < L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{k}\)

You are given that \(L = \frac{{{\pi ^2}}}{6}\).

By taking *k *= 4 , use the upper bound and lower bound for *L *to find an upper bound and lower bound for \(\pi \) . Give your bounds to three significant figures.

**Answer/Explanation**

## Markscheme

\(\mathop {\lim }\limits_{H \to \infty } \int_a^H {\frac{1}{{{x^2}}}{\text{d}}x} = \mathop {\lim }\limits_{H \to \infty } \left[ {\frac{{ – 1}}{x}} \right]_a^H\) *A1*

\(\mathop {\lim }\limits_{H \to \infty } \left( {\frac{{ – 1}}{H} + \frac{1}{a}} \right)\) *A1*

\( = \frac{1}{a}\) *A1*

*[3 marks]*

as \(\left\{ {\frac{1}{{{n^2}}}} \right\}\) is a positive decreasing sequence we consider the function \(\frac{1}{{{x^2}}}\)

we look at \(\int_1^\infty {\frac{1}{{{x^2}}}} {\text{d}}x\) *M1*

\(\int_1^\infty {\frac{1}{{{x^2}}}} {\text{d}}x = 1\) **A1**

since this is finite (allow “limit exists” or equivalent statement) *R1*

\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) converges **AG**

*[3 marks]*

(i)

attempt to shade rectangles *M1*

correct start and finish points for rectangles *A1*

since the area shaded is less that the area of the required staircase we have *R1*

\(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_{k + 1}^\infty {\frac{1}{{{x^2}}}} {\text{d}}x < L\) *AG*

(ii)

attempt to shade rectangles *M1*

correct start and finish points for rectangles *A1*

since the area shaded is greater that the area of the required staircase we have *R1*

\(L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_k^\infty {\frac{1}{{{x^2}}}} {\text{d}}x\) **AG**

**Note: **Alternative shading and rearranging of the inequality is acceptable.

*[6 marks]*

\(\int_{k + 1}^\infty {\frac{1}{{{x^2}}}} {\text{d}}x = \frac{1}{{k + 1}},{\text{ }}\int_k^\infty {\frac{1}{{{x^2}}}} {\text{d}}x = \frac{1}{k}\) **A1A1**

\(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{{k + 1}} < L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{k}\) **AG**

*[2 marks]*

\(\frac{{205}}{{144}} + \frac{1}{5} < \frac{{{\pi ^2}}}{6} < \frac{{205}}{{144}} + \frac{1}{4}{\text{ }}\left( {1.6236… < \frac{{{\pi ^2}}}{6} < 1.6736…} \right)\) *A1*

\(\sqrt {6\left( {\frac{{205}}{{144}} + \frac{1}{5}} \right)} < \pi < \sqrt {6\left( {\frac{{205}}{{144}} + \frac{1}{4}} \right)} \) *(M1)*

\(3.12 < \pi < 3.17\) *A1 N2*

*[3 marks]*

## Examiners report

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

In part (b) the calculation of the integral as equal to 1 only scored 2 of the 3 marks. The final mark was for stating that ‘because the value of the integral is finite (or ‘the limit exists’ or an equivalent statement) then the series converges. Quite a few candidates left out this phrase.

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

Candidates found part (c) difficult. Very few drew the correct series of rectangles and some clearly had no idea of what was expected of them.

Though part (e) could be done without doing any of the previous parts of the question many students were probably put off by the notation because only a minority attempted it.

## Question

Use the limit comparison test to prove that \(\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}} \) converges.

Using the Maclaurin series for \(\ln (1 + x)\) , show that the Maclaurin series for \(\left( {1 + x} \right)\ln \left( {1 + x} \right)\) is \(x + \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^{n + 1}}{x^{n + 1}}}}{{n(n + 1)}}} \).

**Answer/Explanation**

## Markscheme

apply the limit comparison test with \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) *M1*

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{n(n + 1)}}}}{{\frac{1}{{{n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{n(n + 1)}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{n}}} = 1\) **M1A1**

(since the limit is finite and \( \ne 0\) ) both series do the same *R1*

we know that \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) converges and hence \(\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}} \) also converges *R1AG*

*[5 marks]*

\((1 + x)\ln (1 + x) = (1 + x)\left( {x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} – \frac{{{x^4}}}{4}…} \right)\) *A1*

\( = \left( {x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} – \frac{{{x^4}}}{4}…} \right) + \left( {{x^2} – \frac{{x3}}{2} + \frac{{{x^4}}}{3} – \frac{{{x^5}}}{4}…} \right)\)

**EITHER**

\( = x + \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^n}{x^{n + 1}}}}{{n + 1}} + \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^{n + 1}}{x^{n + 1}}}}{n}} } \) **A1**

\( = x + \sum\limits_{n = 1}^\infty {{{( – 1)}^{n + 1}}{x^{n + 1}}\left( {\frac{{ – 1}}{{n + 1}} + \frac{1}{n}} \right)} \) **M1**

**OR**

\(x + \left( {1 – \frac{1}{2}} \right){x^2} – \left( {\frac{1}{2} – \frac{1}{3}} \right){x^3} + \left( {\frac{1}{3} – \frac{1}{4}} \right){x^4} – …\) *A1*

\( = x + \sum\limits_{n = 1}^\infty {{{( – 1)}^{n + 1}}{x^{n + 1}}\left( {\frac{1}{n} – \frac{1}{{n + 1}}} \right)} \) **M1**

\( = x + \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^{n + 1}}{x^{n + 1}}}}{{n(n + 1)}}} \) **AG**

*[3 marks]*

## Examiners report

Candidates and teachers need to be aware that the Limit comparison test is distinct from the comparison test. Quite a number of candidates lost most of the marks for this part by doing the wrong test.

Some candidates failed to state that because the result was finite and not equal to zero then the two series converge or diverge together. Others forgot to state, with a reason, that \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) converges.

Candidates and teachers need to be aware that the Limit comparison test is distinct from the comparison test. Quite a number of candidates lost most of the marks for this part by doing the wrong test.

Some candidates failed to state that because the result was finite and not equal to zero then the two series converge or diverge together. Others forgot to state, with a reason, that \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) converges.

In part (b) finding the partial fractions was well done. The second part involving the use of telescoping series was less well done, and students were clearly not as familiar with this technique as with some others.

Part (c) was the least well done of all the questions. It was expected that students would use explicitly the result from the first part of 4(b) or show it once again in order to give a complete answer to this question, rather than just assuming that a pattern spotted in the first few terms would continue.

Candidates need to be informed that unless specifically told otherwise they may use without proof any of the Maclaurin expansions given in the Information Booklet. There were many candidates who lost time and gained no marks by trying to derive the expansion for \(\ln (1 + x)\).

## Question

Consider the infinite series \(\sum\limits_{n = 1}^\infty {\frac{2}{{{n^2} + 3n}}} \).

Use a comparison test to show that the series converges.

**Answer/Explanation**

## Markscheme

**EITHER**

\(\sum\limits_{n = 1}^\infty {\frac{2}{{{n^2} + 3n}}} < \sum\limits_{n = 1}^\infty {\frac{2}{{{n^2}}}} \) *M1*

which is convergent *A1*

the given series is therefore convergent using the comparison test *AG*

**OR**

\(\mathop {{\text{lim}}}\limits_{n \to \infty } \frac{{\frac{2}{{{n^2} + 3n}}}}{{\frac{1}{{{n^2}}}}} = 2\) *M1A1*

the given series is therefore convergent using the limit comparison test *AG*

*[2 marks]*

## Examiners report

Most candidates were able to answer part (a) and many gained a fully correct answer. A number of candidates ignored the factor 2 in the numerator and this led to candidates being penalised. In some cases candidates were not able to identify an appropriate series to compare with. Most candidates used the Comparison test rather than the Limit comparison test.

## Question

Show that the series \(\sum\limits_{n = 2}^\infty {\frac{1}{{{n^2}\ln n}}} \) converges.

(i) Show that \(\ln (n) + \ln \left( {1 + \frac{1}{n}} \right) = \ln (n + 1)\).

(ii) Using this result, show that an application of the ratio test fails to determine whether or not \(\sum\limits_{n = 2}^\infty {\frac{1}{{n\ln n}}} \) converges.

(i) State why the integral test can be used to determine the convergence or divergence of \(\sum\limits_{n = 2}^\infty {\frac{1}{{n\ln n}}} \).

(ii) Hence determine the convergence or divergence of \(\sum\limits_{n = 2}^\infty {\frac{1}{{n\ln n}}} \).

**Answer/Explanation**

## Markscheme

**METHOD 1**

\((0 < )\frac{1}{{{n^2}\ln (n)}} < \frac{1}{{{n^2}}},{\text{ }}({\text{for }}n \ge 3)\) *A1*

\(\sum\limits_{n = 2}^\infty {\frac{1}{{{n^2}}}} \) converges *A1*

by the comparison test ( \(\sum\limits_{n = 2}^\infty {\frac{1}{{{n^2}}}} \) converges implies) \(\sum\limits_{n = 2}^\infty {\frac{1}{{{n^2}\ln (n)}}} \) converges *R1*

**Note:** Mention of using the comparison test may have come earlier.

Only award ** R1** if previous 2

**s have been awarded.**

*A1***METHOD 2**

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\frac{1}{{{n^2}\ln n}}}}{{\frac{1}{{{n^2}}}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\ln n}} = 0\) *A1*

\(\sum\limits_{n = 2}^\infty {\frac{1}{{{n^2}}}} \) converges *A1*

by the limit comparison test (if the limit is \(0\) and the series represented by the denominator converges, then so does the series represented by the numerator, hence) \(\sum\limits_{n = 2}^\infty {\frac{1}{{{n^2}\ln (n)}}} \) converges *R1*

**Note:** Mention of using the limit comparison test may come earlier.

Do not award the ** R1** if incorrect justifications are given, for example the series “converge or diverge together”.

Only award ** R1** if previous 2

**s have been awarded.**

*A1***[3 marks]**

(i) **EITHER**

\(\ln (n) + \ln \left( {1 + \frac{1}{n}} \right) = \ln \left( {n\left( {1 + \frac{1}{n}} \right)} \right)\) *A1*

**OR**

\(\ln (n) + \ln \left( {1 + \frac{1}{n}} \right) = \ln (n) + \ln \left( {\frac{{n + 1}}{n}} \right)\)

\( = \ln (n) + \ln (n + 1) – \ln (n)\) *A1*

**THEN**

\( = \ln (n + 1)\) *AG*

(ii) attempt to use the ratio test \(\frac{n}{{(n + 1)}}\frac{{\ln (n)}}{{\ln (n + 1)}}\) *M1*

\(\frac{n}{{n + 1}} \to 1{\text{ as }}n \to \infty \) *(A1)*

\(\frac{{\ln (n)}}{{\ln (n + 1)}} = \frac{{\ln (n)}}{{\ln (n) + \ln \left( {1 + \frac{1}{n}} \right)}}\) *M1*

\( \to 1\;\;\;({\text{as }}n \to \infty )\) *(A1)*

\(\frac{n}{{(n + 1)}}\frac{{\ln (n)}}{{\ln (n + 1)}} \to 1\;\;\;({\text{as }}n \to \infty )\;\;\;\)hence ratio test is inconclusive *R1*

**Note:** A link with the limit equalling \(1\) and the result being inconclusive needs to be given for ** R1**.

**[6 marks]**

(i) consider \(f(x) = \frac{1}{{x\ln x}}\;\;\;({\text{for }}x > 1)\) *A1*

\(f(x)\) is continuous and positive *A1*

and is (monotonically) decreasing *A1*

**Note:** If a candidate uses \(n\) rather than \(x\), award as follows

\(\frac{1}{{n\ln n}}\) is positive and decreasing *A1A1*

\(\frac{1}{{n\ln n}}\) is continuous for \(n \in \mathbb{R},{\text{ }}n > 1\) ** A1** (only award this mark if the domain has been explicitly changed).

(ii) consider \(\int_2^R {\frac{1}{{x\ln x}}{\text{d}}x} \) *M1*

\( = \left[ {\ln (\ln x)} \right]_2^R\) *(M1)A1*

\( \to \infty {\text{ as }}R \to \infty \) *R1*

hence series diverges *A1*

**Note:** Condone the use of \(\infty \) in place of \(R\).

**Note:** If the lower limit is not equal to \(2\), but the expression is integrated correctly award ** M0M1A1R0A0**.

**[8 marks]**

**Total [17 marks]**

## Examiners report

In this part the required test was not given in the question. This led to some students attempting inappropriate methods. When using the comparison or limit comparision test many candidates wrote the incorrect statement \(\frac{1}{{{n^2}}}\) converges, (*p*-series) rather than the correct one with \(\sum {} \). This perhaps indicates a lack of understanding of the concepts involved.

There were many good, well argued answers to this part. Most candidates recognised the importance of the result in part (i) to find the limit in part (ii). Generally a standard result such as \(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{n + 1}}} \right) = 1\) can simply be quoted, but other limits such as \(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\ln n}}{{\ln (n + 1)}}} \right) = 1\) need to be carefully justified.

(i) Candidates need to be aware of the necessary conditions for all the series tests.

(ii) The integration was well done by the candidates. Most also made the correct link between the integral being undefined and the series diverging. In this question it was not necessary to initially take a finite upper limit and the use of \(\infty \) was acceptable. This was due to the command term being ‘determine’. In q4b a finite upper limit was required, as the command term was ‘show’. To ensure full marks are always awarded candidates should err on the side of caution and always use limit notation when working out indefinite integrals.

## Question

Consider the infinite series \(S = \sum\limits_{n = 0}^\infty {{u_n}} \) where \({u_n} = \int_{nx}^{(n + 1)\pi } {\frac{{\sin t}}{t}{\text{d}}t} \).

Explain why the series is alternating.

(i) Use the substitution \(T = t – \pi \) in the expression for \({u_{n + 1}}\) to show that \(\left| {{u_{n + 1}}} \right| < \left| {{u_n}} \right|\).

(ii) Show that the series is convergent.

Show that \(S < 1.65\).

**Answer/Explanation**

## Markscheme

as \(t\) moves through the intervals \([0,{\text{ }}\pi ],{\text{ }}[\pi ,{\text{ }}2\pi ],{\text{ }}[2\pi ,{\text{ }}3\pi ],{\text{ }}[3\pi ,{\text{ }}4\pi ]\), *etc*, the sign of \(\sin t\), (and therefore the sign of the integral) alternates \( + ,{\text{ }} – ,{\text{ }} + ,{\text{ }} – \), *etc*, so that the series is alternating *R1*

**Note: **Award ** R1 **only if it includes a clear reason that justifies that the sign of the integrand alternates between − and + and this pattern is valid for all the terms.

The change of signs can be justified by a labelled graph of \(y = \sin (x)\) or \(y = \frac{{\sin x}}{x}\) that shows the intervals \([0,{\text{ }}\pi ],{\text{ }}[\pi ,{\text{ }}2\pi ],{\text{ }}[2\pi ,{\text{ }}3\pi ],{\text{ }} \ldots \)

*[1 mark]*

(i) \({u_{n + 1}} = \int_{(n + 1)\pi }^{(n + 2)\pi } {\frac{{\sin t}}{t}{\text{d}}t} \)

*(M1)*

put \(T = t–\pi \) and \({\text{d}}T = {\text{d}}t\) *(M1)*

the limits change to \(n\pi ,{\text{ }}(n + 1)\pi \)

\(\left| {{u_{n + 1}}} \right| = \int_{n\pi }^{(n + 1)\pi } {\frac{{\left| {\sin (T + \pi )} \right|}}{{T + \pi }}{\text{d}}T} \) (or equivalent) *A1*

\(\left| {\sin (T + \pi )} \right| = \left| {\sin (T)} \right|\) or \(\sin (T + \pi ) = – \sin (T)\) *(M1)*

\( = \int_{n\pi }^{(n + 1)\pi } {\frac{{\left| {\sin T} \right|}}{{T + \pi }}{\text{d}}T} \)

\( < \int_{n\pi }^{(n + 1)\pi } {\frac{{\left| {\sin T} \right|}}{T}{\text{d}}T = \left| {{u_n}} \right|} \) *A1AG*

(ii) \(\left| {{u_n}} \right| = \int_{n\pi }^{(n + 1)\pi } {\frac{{\sin t}}{t}{\text{d}}t} \)

\( < \int_{n\pi }^{(n + 1)\pi } {\frac{1}{t}{\text{d}}t} \) *M1*

\( = [\ln t]_{n\pi }^{(n + 1)\pi }\) *A1*

\( = \ln \left( {\frac{{n + 1}}{n}} \right)\) *A1*

\( \to \ln 1 = 0\) as \(n \to \infty \)

from part (i) \(\left| {{u_n}} \right|\) is a decreasing sequence and since \(\mathop {\lim }\limits_{n \to \infty } \left| {{u_n}} \right| = 0\), *R1*

the series is convergent *AG*

*[9 marks]*

attempt to calculate the partial sums \(\sum\limits_{i = 0}^{n – 1} {{u_i} = \int_0^{n\pi } {\frac{{\sin t}}{t}{\text{d}}t} } \) *(M1)*

the first partial sums are

two consecutive partial sums for \(n \geqslant 4\) *A1A1*

(*eg* \({S_4} = 1.49\) and \({S_5} = 1.63\) or \({S_{100}} = 1.567 \ldots \) and \({S_{101}} = 1.573 \ldots \))

**Note: **These answers must be given to a minimum of 3 significant figures.

the sum to infinity lies between any two consecutive partial sums,

*eg *between 1.49 and 1.63 *R1*

so that \(S < 1.65\) *AG*

**Note: **Award ** A1A1R1 **to candidates who calculate at least two partial sums for only odd values of \(n\) and state that the upper bound is less than these values.

*[4 marks]*

## Examiners report

Very few candidates presented a valid reason to justify the alternating nature of the series. In most cases candidates just reformulated the wording of the question by saying that it changed signs and completely ignored the interval over which the expression had to be integrated to obtain each term.

(i) Most candidates achieved 1 or 2 marks for attempting the given substitution; in most cases candidates failed to find the correct limits of integration for the new variable and then relate the expressions of the consecutive terms of the series. In part (ii) very few correct attempts were seen; in some cases candidates did recognize the conditions for the alternating series to be convergent but very few got close to establish that the limit of the general term was zero.

A few good attempts to use partial sums were seen although once again candidates showed difficulties in identifying what was needed to show the given answer. In most cases candidates just verified with GDC that in fact for high values of *n *the series was indeed less than the upper bound given but could not provide a valid argument that justified the given statement.

## Question

Consider the infinite spiral of right angle triangles as shown in the following diagram.

The \(n{\text{th}}\) triangle in the spiral has central angle \({\theta _n}\), hypotenuse of length \({a_n}\) and opposite side of length 1, as shown in the diagram. The first right angle triangle is isosceles with the two equal sides being of length 1.

Consider the series \(\sum\limits_{n = 1}^\infty {{\theta _n}} \).

Using l’Hôpital’s rule, find \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)\).

(i) Find \({a_1}\) and \({a_2}\) and hence write down an expression for \({a_n}\).

(ii) Show that \({\theta _n} = \arcsin \frac{1}{{\sqrt {(n + 1)} }}\).

Using a suitable test, determine whether this series converges or diverges.

**Answer/Explanation**

## Markscheme

\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)\) is of the form \(\frac{0}{0}\)

and so will equal the limit of \(\frac{{\frac{{\frac{{ – 1}}{2}{{(x + 1)}^{ – \frac{3}{2}}}}}{{\sqrt {1 – \left( {\frac{1}{{x + 1}}} \right)} }}}}{{\frac{{ – 1}}{2}{x^{ – \frac{3}{2}}}}}\) *M1M1A1A1*

**Note: M1 **for attempting differentiation of the top and bottom,

**for derivative of top (only award**

*M1A1***if chain rule is used),**

*M1***for derivative of bottom.**

*A1*\( = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\frac{x}{{(x + 1)}}} \right)}^{\frac{3}{2}}}}}{{\sqrt {\frac{x}{{x + 1}}} }} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{x}{{x + 1}}} \right)\) *M1*

**Note: **Accept any intermediate tidying up of correct derivative for the method mark.

\( = 1\) *A1*

*[6 marks]*

(i) \({a_1} = \sqrt 2 ,{\text{ }}{a_2} = \sqrt 3 \) *A1*

\({a_n} = \sqrt {n + 1} \) *A1*

(ii) \(\sin {\theta _n} = \frac{1}{{{a_n}}} = \frac{1}{{\sqrt {n + 1} }}\) *A1*

**Note: **Allow \({\theta _n} = \arcsin \left( {\frac{1}{{{a_n}}}} \right)\) if \({a_n} = \sqrt {n + 1} \) in b(i).

so \({\theta _n} = \arcsin \frac{1}{{\sqrt {(n + 1)} }}\) *AG*

*[3 marks]*

for \(\sum\limits_{n = 1}^\infty {\arcsin \frac{1}{{\sqrt {(n + 1)} }}} \) apply the limit comparison test (since both series of positive terms) *M1*

with \(\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt n }}} \) *A1*

from (a) \(\mathop {\lim }\limits_{n \to \infty } \frac{{\arcsin \frac{1}{{\sqrt {(n + 1)} }}}}{{\frac{1}{{\sqrt n }}}} = 1\), so the two series either both converge or both diverge *M1R1*

\(\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt 2 }}} \) diverges (as is a \(p\)-series with \(p = \frac{1}{2}\)) *A1*

hence \(\sum\limits_{n = 1}^\infty {{\theta _n}} \) diverges *A1*

*[6 marks]*

## Examiners report

[N/A]

[N/A]

[N/A]

## Question

Given that \(n > {\text{ln}}\,n\) for \(n > 0\), use the comparison test to show that the series \(\sum\limits_{n = 0}^\infty {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) is divergent.

Find the interval of convergence for \(\sum\limits_{n = 0}^\infty {\frac{{{{\left( {3x} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \).

**Answer/Explanation**

## Markscheme

**METHOD 1**

\({\text{ln}}\left( {n + 2} \right) < n + 2\) **(A1)**

\( \Rightarrow \frac{1}{{{\text{ln}}\left( {n + 2} \right)}} > \frac{1}{{n + 2}}\) (for \(n \geqslant 0\)) **A1**

**Note:** Award A0 for statements such as \(\sum\limits_{n = 0}^\infty {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} > \sum\limits_{n = 0}^\infty {\frac{1}{{n + 2}}} \). However condone such a statement if the above * A1 *has already been awarded.

\(\sum\limits_{n = 0}^\infty {\frac{1}{{n + 2}}} \) (is a harmonic series which) diverges **R1**

**Note:** The * R1 *is independent of the

*s.*

**A1**Award * R0 *for statements such as “\(\frac{1}{{n + 2}}\) diverges”.

so \(\sum\limits_{n = 0}^\infty {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) diverges by the comparison test **AG**

**METHOD 2**

\(\frac{1}{{{\text{ln}}\,n}} > \frac{1}{n}\) (for \(n \geqslant 2\)) **A1**

**Note:** Award * A0* for statements such as \(\sum\limits_{n = 2}^\infty {\frac{1}{{{\text{ln}}\,n}}} > \sum\limits_{n = 2}^\infty {\frac{1}{n}} \). However condone such a statement if the above

*has already been awarded.*

**A1**a correct statement linking \(n\) and \(n + 2\) *eg*,

\(\sum\limits_{n = 0}^\infty {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} = \sum\limits_{n = 2}^\infty {\frac{1}{{{\text{ln}}\,n}}} \) or \(\sum\limits_{n = 0}^\infty {\frac{1}{{n + 2}}} = \sum\limits_{n = 2}^\infty {\frac{1}{n}} \) **A1**

**Note:** Award * A0 *for \(\sum\limits_{n = 0}^\infty {\frac{1}{n}} \)

\(\sum\limits_{n = 2}^\infty {\frac{1}{n}} \) (is a harmonic series which) diverges

(which implies that \(\sum\limits_{n = 2}^\infty {\frac{1}{{{\text{ln}}\,n}}} \) diverges by the comparison test) **R1**

**Note:** The * R1 *is independent of the

*s.*

**A1**Award * R0 *for statements such as \(\sum\limits_{n = 0}^\infty {\frac{1}{n}} \) deiverges and “\({\frac{1}{n}}\) diverges”.

Award * A1A0R1 *for arguments based on \(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \).

so \(\sum\limits_{n = 0}^\infty {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) diverges by the comparison test **AG**

**[3 marks]**

applying the ratio test \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left| {\frac{{{{\left( {3x} \right)}^{n + 1}}}}{{{\text{ln}}\left( {n + 3} \right)}} \times \frac{{{\text{ln}}\left( {n + 2} \right)}}{{{{\left( {3x} \right)}^n}}}} \right|\) **M1**

\( = \left| {3x} \right|\) (as \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left| {\frac{{{\text{ln}}\left( {n + 2} \right)}}{{{\text{ln}}\left( {n + 3} \right)}}} \right| = 1\) **A1**

**Note:** Condone the absence of limits and modulus signs.

**Note:** Award * M1A0 *for \(3{x^n}\). Subsequent marks can be awarded.

series converges for \( – \frac{1}{3} < x < \frac{1}{3}\)

considering \(x = – \frac{1}{3}\) and \(x = \frac{1}{3}\) **M1**

**Note:** Award * M1 *to candidates who consider one endpoint.

when \(x = \frac{1}{3}\), series is \(\sum\limits_{n = 0}^\infty {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) which is divergent (from (a)) **A1**

**Note:** Award this * A1 *if \(\sum\limits_{n = 0}^\infty {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) is not stated but reference to part (a) is.

when \(x = – \frac{1}{3}\), series is \(\sum\limits_{n = 0}^\infty {\frac{{{{\left( { – 1} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \) **A1**

\(\sum\limits_{n = 0}^\infty {\frac{{{{\left( { – 1} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \) converges (conditionally) by the alternating series test **R1**

(strictly alternating, \(\left| {{u_n}} \right| > \left| {{u_{n + 1}}} \right|\) for \(n \geqslant 0\) and \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left( {{u_n}} \right) = 0\))

so the interval of convergence of *S* is \( – \frac{1}{3} \leqslant x < \frac{1}{3}\) **A1**

**Note:** The final * A1 *is dependent on previous

*s –*

**A1***ie*, considering correct series when \(x = – \frac{1}{3}\) and \(x = \frac{1}{3}\) and on the final

*.*

**R1**Award as above to candidates who firstly consider \(x = – \frac{1}{3}\) and then state conditional convergence implies divergence at \(x = \frac{1}{3}\).

**[7 marks]**

## Examiners report

[N/A]

[N/A]

## Question

Consider the infinite series \(\sum\limits_{n = 1}^\infty {\frac{{(n – 1){x^n}}}{{{n^2} \times {2^n}}}} \) .

Find the radius of convergence.

Find the interval of convergence.

**Answer/Explanation**

## Markscheme

using the ratio test, \(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{n{x^{n + 1}}}}{{{{(n + 1)}^2}{2^{n + 1}}}} \times \frac{{{n^2}{2^n}}}{{(n – 1){x^n}}}\) *M1*

\( = \frac{{{n^3}}}{{{{(n + 1)}^2}(n – 1)}} \times \frac{x}{2}\) *A1*

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{x}{2}\) *A1*

the radius of convergence *R* satisfies

\(\frac{R}{2} = 1\) so *R* = 2 *A1*

*[4 marks]*

considering *x* = 2 for which the series is

\(\sum\limits_{n = 1}^\infty {\frac{{(n – 1)}}{{{n^2}}}} \)

using the limit comparison test with the harmonic series *M1*

\(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \), which diverges

consider

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n – 1}}{n} = 1\) *A1*

the series is therefore divergent for *x* = 2 *A1*

when *x* = –2 , the series is

\(\sum\limits_{n = 1}^\infty {\frac{{(n – 1)}}{{{n^2}}} \times {{( – 1)}^n}} \)

this is an alternating series in which the \({n^{{\text{th}}}}\) term tends to 0 as \(n \to \infty \) *A1*

consider \(f(x) = \frac{{x – 1}}{{{x^2}}}\) *M1*

\(f'(x) = \frac{{2 – x}}{{{x^3}}}\) *A1*

this is negative for \(x > 2\) so the sequence \(\{ |{u_n}|\} \) is eventually decreasing *R1*

the series therefore converges when *x* = –2 by the alternating series test *R1*

the interval of convergence is therefore [–2, 2[ *A1*

*[9 marks]*

## Examiners report

[N/A]

[N/A]