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Playing around with Mathematica, I found a rather nice result for this summation \begin{align} \sum_{n=0}^\infty \frac{1}{(n^2 + a)(n^2 +b) (n^2 + c)} &= \frac{1}{2 a b c} - \frac{ \pi \left[ \sqrt{ab}(a-b) \coth(\pi\sqrt{c}) + \sqrt{bc}(b-c) \coth(\pi \sqrt{a}) + \sqrt{ca}(c-a) \coth( \pi \sqrt{b}) \right]}{2 \sqrt{a b c} \, (a-b)(b-c)(c-a)}. \end{align}

Does anybody know how to obtain this result analytically?

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Partial fraction decomposition gives $$\frac1{(n^2+a)(n^2+b)(n^2+c)}=\sum_{\mathrm{cyc}}\frac1{(a-c)(b-c)}\cdot\frac1{n^2+c}$$ Now use the well-known sum $$\sum_{n=0}^\infty\frac1{n^2+k}=\frac{1+\sqrt k\pi\coth\sqrt k\pi}{2k}$$ which should lead you to the final result.

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  • $\begingroup$ Could you please elaborate on the partial fraction you mention? What you've written is degree $6$ in $n$ (in denominator) on the l.h.s but the r.h.s seems to be of degree $2$ $\endgroup$
    – Saïd M
    Commented Oct 12, 2020 at 10:13
  • $\begingroup$ @SaSoMo PFD reduces the seemingly degree-$6$ term into a sum of three degree-$2$ terms, which are then evaluated separately. $\endgroup$
    – Parcly Taxel
    Commented Oct 12, 2020 at 10:15

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