Use the limit comparison test to show that the series $\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2} + 2}}}$ is convergent.

Find the interval of convergence for $S$.

$\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{{n^2} + 2}}}}{{\frac{1}{{{n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{{n^2} + 2}} = \left( {\mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{2}{{{n^2} + 2}}} \right)} \right)$     M1

$= 1$     A1

since $\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}}$ converges (a $p$-series with $p = 2$)     R1

by limit comparison test, $\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2} + 2}}}$ also converges     AG

Notes:     The R1 is independent of the A1.

[3 marks]

applying the ratio test $\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{(x - 3)}^{n + 1}}}}{{{{(n + 1)}^2} + 2}} \times \frac{{{n^2} + 2}}{{{{(x - 3)}^n}}}} \right|$     M1A1

$= \left| {x - 3} \right|{\text{ }}\left( {{\text{as }}\mathop {\lim }\limits_{n \to \infty } \frac{{({n^2} + 2)}}{{{{(n + 1)}^2} + 2}} = 1} \right)$     A1

converges if $\left| {x - 3} \right| < 1$ (converges for $2 < x < 4$)     M1

considering endpoints $x = 2$ and $x = 4$     M1

when $x = 4$, series is $\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2} + 2}}}$, convergent from (a)     A1

when $x = 2$, series is $\sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}}}{{{n^2} + 2}}}$     A1

EITHER

$\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2} + 2}}}$ is convergent therefore $\sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}}}{{{n^2} + 2}}}$ is (absolutely) convergent     R1

OR

$\frac{1}{{{n^2} + 2}}$ is a decreasing sequence and $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2} + 2}} = 0$ so series converges by the alternating series test     R1

THEN

interval of convergence is $2 \leqslant x \leqslant 4$     A1

Note:     The final A1 is dependent on previous A1s – ie, considering correct series when $x = 2$ and $x = 4$ and on the final R1.

[9 marks]