Let \(\varpi=\dfrac{\Gamma^2\left(\frac14\right)}{2\sqrt{2\pi}}=2.62205755\ldots\) be the lemniscate constant. Then,
\[\Large\displaystyle\sum_{n=1}^\infty\dfrac{1}{\sinh^4(\pi n)}=\dfrac{\varpi^4}{30\pi^4}+\dfrac{1}{3\pi}-\dfrac{11}{90}\]

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An interesting infinite product!

\[\displaystyle\prod_{n=1}^\infty\coth^{(-1)^n}\left(\dfrac{\pi n}{2}\right)=\dfrac{\sqrt\pi}{\sqrt[4]2\sqrt{\varpi}}\]

where \(\varpi=2.62205755\ldots\) is the lemniscate constant (the ratio of the perimeter of Bernoulli's lemniscate to its diameter, analogous to the definition of \(\pi\) for the circle). Both \(\varpi\) and \(\pi\) are proven to be transcendental.

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