Is it possible to eliminate the inner sum to evaluate numerically?

Question

Any hints on how to simplify the following double sum to be able to find the sum at least numerically?

$$\sum_{n=2}^{\infty}\frac1{n(n^2-1)} \sum_{k=1}^\infty \frac{(k-1/n)^{2n-2}}{(k+1/n)^{2n+2}}$$

Series comes from quantum mechanics (2nd order perturbation theory).

Answer 1

Numerically the sum isn't terribly easy: it behaves as $$ \sum_{n\geq 2, k\geq 1} \frac{1}{n(n^2-1)} \frac{e^{-4/k}}{(k^2-n^{-2})^2}, $$ so it converges linearly. Most numerical methods that can be expected to do well only really work for either alternating, or quickly converging sums. With a sum that converges linearly, you can only really get anywhere if you can guess what the remainder looks like. In your case, the remainder behaves as $k^{-3}$ or $n^{-2}$, so it's quite regular and there are Richardson, Shanks and Levin methods to try first.

I tried it in mpmath (part of sage) in python, and I got $$\begin{array}{rr} &0.00920\ 70258\ 69569\ 42996\ 92855\ 67863\ 59226\ 74177\\ &52180\ 49145\ 07649\ 14972\ 00277\ 81544\ 60503\ 04938\\ &73843\ 97518\ 00620\ 22842\ 37461\ 13380\ 72993\ 00261\\ &71196\ 37097\ 25227\ 29413\ 59023\ 59030\ 36624\ 51643\\ &60770\ 76310\ 65123\ 95919\ 68631\ 77487\ 86050\ 33349 \end{array}$$ using the levin method, using this code:

nsum((lambda n, k: (k-1/n)**(2*n-2)/((k+1/n)**(2*n+2)*n*(n*n-1))), [2,inf], [1,inf], method='levin', verbose=True)

EDIT In Mathematica, this seems to work, but doesn't give nearly as many digits, only about six (four correct):

NSum[1/(n (n^2 - 1))*(k - 1/n)^(2 n - 2)/(k + 1/n)^(2 n + 2), {k, 1, \[Infinity]}, {n, 2, \[Infinity]}, Method -> "WynnEpsilon"]

References that I used myself: Numerical Recipes book, http://dlmf.nist.gov/3.9 (clearest, shortest explanation), original papers for the methods (see references in the DLMF link). Also, just general mathematical considerations: acceleration methods make lots of assumptions about the series' asymptotic behaviour, so those assumptions need to be correct.