# 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!

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.
• 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? – mAlex Jun 26 '14 at 19:32
• Just got it. We should let $s=t^2$ and then everything becomes clearer. – mAlex Jun 26 '14 at 19:35
• I meant $t=s^2$ of course. My bad. – mAlex Jun 26 '14 at 20:12