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

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]

(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]

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.