Consider the infinite series $S = \sum\limits_{n = 1}^\infty  {\frac{{{x^n}}}{{{2^{2n}}\left( {2{n^2} - 1} \right)}}}$.

(a)     Determine the radius of convergence.

(b)     Determine the interval of convergence.

(a) let ${T_n}$ denote the $n{\text{th}}$ term

consider

$\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{{x^{(n + 1)}}}}{{{2^{2(n + 1)}}\left( {2{{[n + 1]}^2} - 1} \right)}} \times \frac{{{2^2}\left( {2{n^2} - 1} \right)}}{{{x^n}}}$     M1

$= \frac{x}{{{2^2}}} \times \frac{{\left( {2{n^2} - 1} \right)}}{{\left( {2{{[n + 1]}^2} - 1} \right)}}$     A1

$\to \frac{x}{4}$ as $n \to \infty$     A1

so the radius of convergence is $4$     A1

[4 marks]

(b)     we need to consider $x =  \pm 4$     R1

$S(4) = \sum\limits_{n = 1}^\infty  {\frac{1}{{\left( {2{n^2} - 1} \right)}}}$     A1

$S(4) < \sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}}$     M1

$\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}}$ is convergent; therefore by the comparison test $S(4)$ is convergent     R1

$S( - 4) = \sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}}}{{\left( {2{n^2} - 1} \right)}}}$     A1

EITHER

this series is convergent because it is absolutely convergent     R1

OR

this series is alternating and is convergent     R1

THEN

the interval of convergence is therefore $\left[ { - 4,{\text{ }}4} \right]$     A1

Note: The final A1 is independent of any of the previous marks.

[7 marks]