Find the radius of convergence of the series.

Hence deduce the interval of convergence.

using the ratio test (and absolute convergence implies convergence)     (M1)

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{{{( - 1)}^{n + 1}}{x^{n + 1}}}}{{(n + 1){2^{n + 1}}}}}}{{\frac{{{{( - 1)}^n}{x^n}}}{{(n){2^n}}}}}} \right|\)     A1A1

Note: Award A1 for numerator, A1 for denominator.

\( = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^{n + 1}} \times {x^{n + 1}} \times n \times {2^n}}}{{{{( - 1)}^n} \times (n + 1) \times {2^{n + 1}} \times {x^n}}}} \right|\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{2(n + 1)}}\left| x \right|\)     (A1)

\( = \frac{{\left| x \right|}}{2}\)     A1

for convergence we require \(\frac{{\left| x \right|}}{2} < 1\)     M1

\( \Rightarrow \left| x \right| < 2\)

hence radius of convergence is 2     A1

[7 marks]

we now need to consider what happens when \(x = \pm 2\)     (M1)

when x = 2 we have \(\sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}}}{n}} \) which is convergent (by the alternating series test)     A1

when x = −2 we have \(\sum\limits_{n = 1}^\infty  {\frac{1}{n}} \) which is divergent     A1

hence interval of convergence is \(] - 2,{\text{ }}2]\)     A1

[4 marks]

Most candidates were able to start (a) and a majority gained a fully correct answer. A number of candidates were careless with using the absolute value sign and with dealing with the negative signs and in the more extreme cases this led to candidates being penalised. Part (b) caused more difficulties, with many candidates appearing to know what to do, but then not succeeding in doing it or in not understanding the significance of the answer gained.

Most candidates were able to start (a) and a majority gained a fully correct answer. A number of candidates were careless with using the absolute value sign and with dealing with the negative signs and in the more extreme cases this led to candidates being penalised. Part (b) caused more difficulties, with many candidates appearing to know what to do, but then not succeeding in doing it or in not understanding the significance of the answer gained.