Primarily, I would like to know what could be done with this series:
$$ \sum_{n=2}^{\infty}\frac{n^3}{(n^2-1)^3}\left(\frac{n-1}{n+1}\right)^{2n}$$
As hardmath says in his comment, the series converges. I tried with Mathematica which gives for the sum 0.00526589. Is it possible to obtain some analytical result/approximation for the sum of the series?
Moreover, I would like to simplify the following sum:
$$ \sum_{n=2}^{\infty}\frac{(n-1)^{2n-3}}{n(n+1)^{2n+1}}\left[ \sum_{m=0}^{n-1}\left(\frac{n+1}{n-1}\right)^m\right]^2$$
Such sums appear in quantum mechanics when dealing with second order perturbation theory (the first mentioned case comes out when dealing with the Coulomb potential 1/r perturturbed by a k^2/r term). Any hints appreciated.
Update: As hardmath says in his comment, in the "second series the finite inner sum (over m) is geometric, so it can be replaced by an explicit expression in terms of n".
This means that we can simplify the double sum into a single one:
$$ \frac1{4}\sum_{n=2}^{\infty}\frac1{n(n^2-1)}\left[1-2\left(\frac{n+1}{n-1}\right)^n+\left(\frac{n+1}{n-1}\right)^{2n}\right] $$
which is closer in form to the first mentioned.