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]