Can you get a closed-form for $\sum_{j=0}^{\infty}\frac{2^{2j-1}B_{2j}}{(2j)!}$?

Question

Let $B_{k}$ the kth Bernoulli number, then using their asymptotic I can justify the absolute convergence of this series

$$\sum_{j=0}^{\infty}\frac{2^{2j-1}B_{2j}}{(2j)!},$$ since, if there are no mistakes $$\lim_{n\to\infty}\frac{c_{n+1}}{c_{n}}=\frac{1}{\pi^2}<1,$$ when $c_n=\frac{2^{2n-1}|B_{2n}|}{(2n)!}$.

Question. Can you get a closed-form for this series $$\sum_{j=0}^{\infty}\frac{2^{2j-1}B_{2j}}{(2j)!}?$$

I would like compute it more precisely than $0.656518$. I've used in Wolfram Alpha this

2^(2n-1) BernoulliB[2n]/(2n)!

but I don't know if I can compute it more precisely.

Thanks in advance.

Answer 1

According to equation $(30)$ in http://mathworld.wolfram.com/BernoulliNumber.html, $$ \frac t2 \coth \frac t2 = \sum_{n=0}^\infty \frac{B_{2n}t^{2n}}{(2n)!} $$ Choosing $t = 2$ gives $$ \sum_{j=0}^{\infty}\frac{2^{2j-1}B_{2j}}{(2j)!} = \frac 12 \coth(1) \approx 0.65651764275 \, . $$

Answer 2

Use the generating function for Bernoulli numbers $$ g(x)={x\over e^x-1}=\sum_{j=0}^\infty B_{j}{x^j\over j!}\\ \implies g(-x)=-{x\over e^{-x}-1}=\sum_{j=0}^\infty (-1)^jB_{j}{x^j\over j!}\\ \implies g(x)+g(-x)={x(e^x+1)\over e^{x}-1}=2\sum_{j=0}^\infty B_{2j}{x^{2j}\over {(2j)}!}\\ \implies\sum_{j=0}^\infty B_{2j}{2^{2j-1}\over {(2j)}!}={1\over2}{\coth(1)} $$