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.

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]

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.