Is there a closed-form of $\sum_{n=0}^{\infty }\frac{|B_n|}{n!}=??$

Question

Is there a closed-form of

$$\sum_{n=0}^{\infty }\frac{|B_n|}{n!}=??$$ where $B_n$ Bernoulli number

Thanks

Answer 1

We have: $$\sum_{n\geq 0}\frac{|B_n|}{n!} = \frac{3}{2}+\sum_{n\geq 1}\frac{|B_{2n}|}{(2n)!}= \frac{3}{2}-\sum_{n\geq 1}\frac{B_{2n}(-1)^n}{(2n)!}.$$ Since: $$ \frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}z^n $$ we have: $$ \frac{z}{2}\coth\frac{z}{2}=\frac{1}{2}\left(\frac{z}{e^z-1}+\frac{-z}{e^{-z}-1}\right) = \sum_{n\geq 0}\frac{B_{2n}}{(2n)!}z^{2n} $$ as well as: $$ \frac{z}{2}\cot\frac{z}{2}=\sum_{n\geq 0}\frac{B_{2n}(-1)^n}{(2n)!}z^{2n}$$ leading to: $$\sum_{n\geq 0}\frac{|B_n|}{n!}=\color{red}{\frac{5-\cot\frac{1}{2}}{2}}.$$

Answer 2

Hint:

use the taylor series of $\cot(x)$

Answer 3

Sure: The sign of $B_{2n}$ is $(-1)^{n+1}$, so use the generating function $$\sum B_n \frac{z^n}{n!} = \frac{z}{e^z - 1},$$ with $z = i$.