Question
a.(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]
b.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]
▶️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
a.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]
b.Show that the series \(\sum\limits_{r = 2}^\infty {\frac{{{{( – 1)}^r}}}{{r\ln r}}} \) is convergent but not absolutely convergent.[5]
▶️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
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
a.Prove by induction that \(n! > {3^n}\), for \(n \ge 7,{\text{ }}n \in \mathbb{Z}\).[5]
b.Hence use the comparison test to prove that the series \(\sum\limits_{r = 1}^\infty {\frac{{{2^r}}}{{r!}}} \) converges.[6]
▶️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]
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]
Examiners report
[N/A]
[N/A]
Question
a.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]
b.Find the interval of convergence for \(\sum\limits_{n = 0}^\infty {\frac{{{{\left( {3x} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \).[7]
▶️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]
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]
Examiners report
[N/A]
[N/A]