Evaluate k = 1^ (-1)^k k^2(2k^2 + 1)^2

Question:

Evaluate{eq}\displaystyle \ \sum_{k = 1}^ \infty \frac {(-1)^k k^2}{(2k^2 + 1)^2} {/eq}

Alternating Series Test:

To check for convergence and divergence of the series {eq}\sum\limits_{n = 1}^\infty {{l_n}} {/eq}, the series can be represented as {eq}{l_n} = {( - 1)^{n + 1}}{k_{n\,\,\,\,\,\,\,}}{\text{or }}{l_n} = {( - 1)^n}{k_{n\,\,\,\,\,}} {/eq}, where {eq}{k_n} \geqslant 0,\forall n,\, {/eq}.

Then for convergence of the series, following conditions must be satisfied;

(1){eq}\mathop {\lim }\limits_{n \to \infty } {k_n} = 0 {/eq}.

(2){eq}\left\{ {{k_n}} \right\} {/eq} is decreasing sequence.

Answer and Explanation:

Given that: {eq}\displaystyle \sum\limits_{k = 1}^\infty {\frac{{{{( - 1)}^k}{k^2}}}{{{{(2{k^2} + 1)}^2}}}} {/eq}

{eq}\displaystyle\ \eqalign{ ...

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