Find the radius of convergence of the infinite series

$$\frac{1}{2}x + \frac{{1 \times 3}}{{2 \times 5}}{x^2} + \frac{{1 \times 3 \times 5}}{{2 \times 5 \times 8}}{x^3} + \frac{{1 \times 3 \times 5 \times 7}}{{2 \times 5 \times 8 \times 11}}{x^4} + \ldots {\text{ .}}$$

Determine whether the series $\sum\limits_{n = 1}^\infty {\sin \left( {\frac{1}{n} + n\pi } \right)}$ is convergent or divergent.

the nth term is

${u_n} = \frac{{1 \times 3 \times 5 \ldots (2n - 1)}}{{2 \times 5 \times 8 \ldots (3n - 1)}}{x^n}$     M1A1

(using the ratio test to test for absolute convergence)

$\frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{(2n + 1)}}{{(3n + 2)}}\left| x \right|$     M1A1

$\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{2\left| x \right|}}{3}$     A1

let R denote the radius of convergence

then $\frac{{2R}}{3} = 1$ so $r = \frac{3}{2}$     M1A1

Note: Do not penalise the absence of absolute value signs.

[7 marks]

using the compound angle formula or a graphical method the series can be written in the form     (M1)

$\sum\limits_{n = 1}^\infty {{u_n}}$ where ${u_n} = {( - 1)^n}\sin \left( {\frac{1}{n}} \right)$     A2

since $\frac{1}{n} < \frac{\pi }{2}$ i.e. an angle in the first quadrant,     R1

it is an alternating series     R1

${u_n} \to 0{\text{ as }}n \to \infty$     R1

and $\left| {{u_{n + 1}}} \right| < \left| {{u_n}} \right|$     R1

it follows that the series is convergent     R1

[8 marks]

Solutions to this question were generally disappointing. In (a), many candidates were unable even to find an expression for the nth term so that they could not apply the ratio test.

Solutions to this question were generally disappointing. In (b), few candidates were able to rewrite the nth term in the form $\sum {{{( - 1)}^n}\sin \left( {\frac{1}{n}} \right)}$ so that most candidates failed to realise that the series was alternating.