<s>[INST] Sum of infinite series $\sum_{n=0}^{\infty}=\frac{n^2}{4n^2-1}t^n$ 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!
 [/INST] 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. 
</s>