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]