IB DP Maths Topic 9.2 Tests for convergence: comparison test; limit comparison test; ratio test; integral test HL Paper 3

 

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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.

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

[8]
a.

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

[11]
b.
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]

a.

(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]

b.

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.

a.

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.

b.

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.

[4]
a.

(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 \(n \in {\mathbb{Z}^ + }\).

[6]
b.

(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 \(n \in {\mathbb{Z}^ + }\).

[4]
c.

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.

[2]
d.
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]

a.

(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]

b.

(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]

c.

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]

d.

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).

a.

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).

b.

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).

c.

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).

d.

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.

[3]
a.

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)     \(\varepsilon  = 0.00001\).

[4]
b.

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.

[6]
c.

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

[2]
d.
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]

a.

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

b.

\({u_n} \to L\) 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]

c.

\({u_n} – L > \frac{7}{{4n}}\) (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]

d.

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. 

a.

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.

b.

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.

c.

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.

d.

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.

[6]
a.

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

[5]
b.
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]

a.

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]

b.

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.

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.

b.

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 \(a{\text{ (where }}a \in {\mathbb{R}^ + })\).

[3]
a.

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

[3]
b.

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

[6]
c.

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

[2]
d.

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.

[3]
e.
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]

a.

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]

b.

(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]

c.

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

d.

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

e.

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)

a.

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.

b.

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.

c.

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)

d.

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.

e.

Question

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

[5]
a.

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

[3]
c.
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]

a.

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

c.

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.

a.

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 \(\ln (1 + x)\).

c.

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.

[3]
a.

(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.

[6]
b.

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

[8]
c.
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 A1s have been awarded.

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 A1s have been awarded.

[3 marks]

a.

(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]

b.

(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]

c.

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.

a.

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.

b.

(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.

c.

Question

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

[3]
a.

(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.

[6]
b.

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

[8]
c.
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 A1s have been awarded.

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 A1s have been awarded.

[3 marks]

a.

(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]

b.

(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]

c.

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.

a.

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.

b.

(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.

c.

Question

Prove by induction that \(n! > {3^n}\), for \(n \ge 7,{\text{ }}n \in \mathbb{Z}\).

[5]
a.

Hence use the comparison test to prove that the series \(\sum\limits_{r = 1}^\infty  {\frac{{{2^r}}}{{r!}}} \) converges.

[6]
b.
Answer/Explanation

Markscheme

if \(n = 7\) then \(7! > {3^7}\)     A1

so true for \(n = 7\)

assume true for \(n = k\)     M1

so \(k! > {3^k}\)

consider \(n = k + 1\)

\((k + 1)! = (k + 1)k!\)     M1

\( > (k + 1){3^k}\)

\( > 3.3k\;\;\;({\text{as }}k > 6)\)     A1

\( = {3^{k + 1}}\)

hence if true for \(n = k\) then also true for \(n = k + 1\). As true for \(n = 7\), so true for all \(n \ge 7\).     R1

Note:     Do not award the R1 if the two M marks have not been awarded.

[5 marks]

a.

consider the series \(\sum\limits_{r = 7}^\infty  {{a_r}} \), where \({a_r} = \frac{{{2^r}}}{{r!}}\)     R1

Note:     Award the R1 for starting at \(r = 7\)

compare to the series \(\sum\limits_{r=7}^\infty  {{b_r}} \) where \({b_r} = \frac{{{2^r}}}{{{3^r}}}\)     M1

\(\sum\limits_{r = 7}^\infty  {{b_r}} \) is an infinite Geometric Series with \(r = \frac{2}{3}\) and hence converges     A1

Note:     Award the A1 even if series starts at \(r = 1\).

as \(r! > {3^r}\) so \((0 < ){a_r} < {b_r}\) for all \(r \ge 7\)     M1R1

as \(\sum\limits_{r = 7}^\infty  {{b_r}} \) converges and \({a_r} < {b_r}\) so \(\sum\limits_{r = 7}^\infty  {{a_r}} \) must converge

Note:     Award the A1 even if series starts at \(r = 1\).

as \(\sum\limits_{r = 1}^6 {{a_r}} \) is finite, so \(\sum\limits_{r = 1}^\infty  {{a_r}} \) must converge     R1

Note:     If the limit comparison test is used award marks to a maximum of R1M1A1M0A0R1.

[6 marks]

Total [11 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

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.

[1]
a.

(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.

[9]
b.

Show that \(S < 1.65\).

[4]
c.
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]

a.

(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]

b.

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]

c.

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.

a.

(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.

b.

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.

c.

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 \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)\).

[6]
a.

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

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

[3]
b.

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

[6]
c.
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, M1A1 for derivative of top (only award M1 if chain rule is used), A1 for derivative of bottom.

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

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

\( = 1\)    A1

[6 marks]

a.

(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]

b.

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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

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.

[3]
a.

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

[7]
b.
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 A1s.

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 A1 has already been awarded.

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 A1s.

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]

a.

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 A1s – 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]

b.

Examiners report

[N/A]

a.

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b.

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