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This problem arose in finding a sum of expectations of Geometric RVs.

Any idea how to simplify the sum

$$\sum_{k=0}^{\frac{n}{2}-1}\frac{1}{(2n^2)^{c'}-(n^2+4k^2)^{c'}}$$

where $c'$ is an integer. It would be good to obtain some digamma function from it.

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Note that $(2n^2)^\ell-(n^2+4k^2)^\ell=(n^2-4k^2)\sum_{k=0}^{\ell-1}(2n^2)^k(n^2+4k^2)^{\e‌​ll-k-1}$... that should help with doing a partial-fraction decomposition of your summand. –  J. M. Apr 14 '11 at 1:12
    
is there an equal sign somewhere in your comment?) –  sigma.z.1980 Apr 14 '11 at 3:40
    
Yes. ${}\qquad$ –  J. M. Apr 14 '11 at 4:14
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