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I have this problem, finding infinite sum of this series:

$$\sum_{n=0}^{\infty}\frac{n^2}{4n^2-1}t^n$$

It should be done using derivatives and integrals, like for example:

$$\sum_{n=1}^{\infty}\frac{t^n}{n}=\sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1}=\sum_{n=0}^{\infty}\int_{0}^{t}s^nds=\int_{0}^{t}\sum_{n=0}^{\infty}s^nds=\int_{0}^{t}\frac{1}{1-s}ds=-ln(1-t)$$

I could think about doing this:

$$\sum_{n=0}^{\infty}\frac{n^2}{4n^2-1}t^n=\sum_{n=0}^{\infty}\frac{n^2}{(2n-1)(2n+1)}t^n=\frac{1}{2}\sum_{n=0}^{\infty}\frac{n^2}{(2n-1)}t^n-\frac{1}{2}\sum_{n=0}^{\infty}\frac{n^2}{(2n+1)}t^n=\ldots$$

but then again, I don't know how could I make it to the end.

Any help would be very appreciated. Thanks!

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You are on the right way. Rewrite the summand as $\frac{1}{4} \cdot \frac{4n^2 -1 + 1}{4 n^2 -1} t^n$ then you get a Geometric series and the second term will be easier - expand it in partial fractions as you did.

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  • $\begingroup$ Okay, that was good hint, and after few steps I get to this part: $$\sum_{n=0}^{\infty}\frac{t^n}{2n-1}$$ which seems pretty easy, but I can't manage to do it properly. Any help on that one? $\endgroup$ – mAlex Jun 26 '14 at 19:32
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    $\begingroup$ Just got it. We should let $s=t^2$ and then everything becomes clearer. $\endgroup$ – mAlex Jun 26 '14 at 19:35
  • $\begingroup$ I meant $t=s^2$ of course. My bad. $\endgroup$ – mAlex Jun 26 '14 at 20:12
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The series \begin{align} S = \sum_{n=0}^{\infty} \frac{ n^{2} \ t^{n} }{ 4n^{2}-1} \end{align} can be reduced as follows. \begin{align} S &= \frac{1}{4} \sum_{n=0}^{\infty} \frac{(4n^{2}-1) + 1}{4n^{2}-1} \ t^{n} \\ &= \frac{1}{4} \sum_{n=0}^{\infty} t^{n} + \frac{1}{8} \sum_{n=0}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \ t^{n} \\ &= \frac{1}{4(1-t)} + \frac{1}{8} \left[ \left( \sqrt{t} + \frac{1}{\sqrt{t}} \right) \tanh^{-1}(\sqrt{t}) - 1 \right], \end{align} where \begin{align} \sum_{n=0}^{\infty} \frac{t^{n}}{2n-1} = \sqrt{t} \tanh^{-1}(\sqrt{t}) -1 \end{align} and \begin{align} \sum_{n=0}^{\infty} \frac{t^{n}}{2n+1} = \frac{1}{\sqrt{t}} \tanh^{-1}(\sqrt{t}) \end{align} was used.

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