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.

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    $\begingroup$ What motivated these Questions? Some context would be helpful to a Reader willing to provide useful information. $\endgroup$
    – hardmath
    Dec 21, 2014 at 3:49
  • $\begingroup$ @hardmath Motivation added:-) $\endgroup$
    – wondering
    Jan 22, 2015 at 12:26
  • $\begingroup$ Thanks. The first series converges (e.g. by comparison with the sum of $1/n^3$), but I suspect you want further information about "what could be done wit[h] this series". Can you be more specific about what is known and what additional analysis would be useful to you? $\endgroup$
    – hardmath
    Jan 22, 2015 at 12:33
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    $\begingroup$ For your first sum, I get 0.00526589019306362497, did you write it correctly? ISC found nothing for 0.00526589019306362497, so perhaps closed form evaluation will not be possible. $\endgroup$
    – GEdgar
    Jan 22, 2015 at 13:32
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    $\begingroup$ In your second series the finite inner sum (over $m$) is geometric, so it can be replaced by an explicit expression in terms of $n$. Have you explored this? $\endgroup$
    – hardmath
    Jan 24, 2015 at 14:14


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