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]