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## infinite series sum

By inspection, we have the following decomposition:

$\displaystyle \frac{1}{(36n^2-1)^2} \equiv \frac{1}{4} \left( \frac{1}{(6n+1)^2}+\frac{1}{(6n-1)^2}+\frac{1}{6n+1}-\frac{1}{6n-1} \right)$

Then the definition of the trigamma and diagamma functions can be applied, combined with integer shifts and reflection formulae to simplify the sums.

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Evaluation of \(\displaystyle \displaystyle \sum^{\infty}_{n=1}\frac{1}{(36n^2-1)^2}\)

$\displaystyle \frac{1}{(36n^2-1)^2} \equiv \frac{1}{4} \left( \frac{1}{(6n+1)^2}+\frac{1}{(6n-1)^2}+\frac{1}{6n+1}-\frac{1}{6n-1} \right)$

Then the definition of the trigamma and diagamma functions can be applied, combined with integer shifts and reflection formulae to simplify the sums.

Thanks paradoxica.

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