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 \(x + \sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^{n + 1}}{x^{n + 1}}}}{{n(n + 1)}}} \).

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]

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