Prove that $\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$ Prove that $$\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$$
My idea is to find the Taylor series of $\frac{1}{(e^x-1)^2}$, but it seems not useful.
Any helps, thanks

We can also use the identity $$x\cot\left(x\right)-1=\sum_{n\geq1}\frac{B_{2n}\left(-4\right)^{n}x^{2n}}{\left(2n\right)!},\,\left|x\right|<\frac{\pi}{2}
 $$ so taking the derivative and manipulating a bit $$x\cot\left(x\right)-\frac{x^{2}}{\sin^{2}\left(x\right)}=\sum_{n\geq1}\frac{B_{2n}\left(-4\right)^{n}x^{2n}}{\left(2n-1\right)!}
 $$ hence taking $x=\frac{i}{2}
 $ we get $$\sum_{n\geq1}\frac{B_{2n}}{\left(2n-1\right)!}=\frac{i\cot\left(i/2\right)}{2}+\frac{1}{4\sin^{2}\left(i/2\right)}=\frac{1}{2}-\frac{1}{\left(e-1\right)^{2}}.$$
