\input amstex  \text{sech}\left(\frac{\sqrt{3}\pi}{2}\right)=\frac{4}{\pi}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{(2k-1)^2+3}=\frac{1}{\pi}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1}