Find the radius of convergence.

Find the interval of convergence.

using the ratio test, \(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{n{x^{n + 1}}}}{{{{(n + 1)}^2}{2^{n + 1}}}} \times \frac{{{n^2}{2^n}}}{{(n - 1){x^n}}}\)     M1

\( = \frac{{{n^3}}}{{{{(n + 1)}^2}(n - 1)}} \times \frac{x}{2}\)     A1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{x}{2}\)     A1

the radius of convergence R satisfies

\(\frac{R}{2} = 1\) so R = 2     A1

[4 marks]

considering x = 2 for which the series is

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

using the limit comparison test with the harmonic series     M1

\(\sum\limits_{n = 1}^\infty  {\frac{1}{n}} \), which diverges

consider

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n - 1}}{n} = 1\)     A1

the series is therefore divergent for x = 2     A1

when x = –2 , the series is

\(\sum\limits_{n = 1}^\infty  {\frac{{(n - 1)}}{{{n^2}}} \times {{( - 1)}^n}} \)

this is an alternating series in which the \({n^{{\text{th}}}}\) term tends to 0 as \(n \to \infty \)     A1

consider \(f(x) = \frac{{x - 1}}{{{x^2}}}\)     M1

\(f'(x) = \frac{{2 - x}}{{{x^3}}}\)     A1

this is negative for \(x > 2\) so the sequence \(\{ |{u_n}|\} \) is eventually decreasing     R1

the series therefore converges when x = –2 by the alternating series test     R1

the interval of convergence is therefore [–2, 2[     A1

[9 marks]