Series involving Bernoulli's numbers

Question

I need to find the exact sum of this series which involves Bernoulli's numbers:

$$\sum_{k=1}^\infty {{B_{2k}(k-1)!\over (2k)!}}$$

It converges very quickly but I'm knew to this kind of problems so I don't have much ideas on what to do.

Answer 1

You might start from the definition of Bernoulli numbers using the generating function $t/(e^t-1)$. We have

$$ \dfrac{(t-2)\; e^t +t+2}{2 (e^t-1)} = -1 + \dfrac{t}{2} + \dfrac{t}{e^t-1} = \sum_{k=1}^\infty \dfrac{B_{2k}}{(2k)!} t^{2k} $$ Now multiply by $2 e^{-t^2}/t$ and integrate from $0$ to $\infty$ (since $\int_0^\infty t^{2k-1} e^{-t^2}\; dt = (k-1)!/2$). Your sum is expressed as an integral:

$$ \int_0^\infty \dfrac{(t-2)\; e^t +t+2}{ t\; (e^t-1)} e^{-t^2}\; dt $$

Unfortunately I don't know if the integral can be done in closed form.

The numerical value is approximately $0.082006067535004$. The Inverse Symbolic Calculator and Maple's "identify" function produced no hits.

Do you have any reason to believe there is a closed form for this?

Answer 2

Since: $$ \sum_{k\geq 0}\frac{B_k}{k!}z^k=\frac{z}{e^z-1}\tag{1}$$ we have: $$ \sum_{k\geq 0}\frac{B_{2k}}{(2k)!}\,z^{2k} = \frac{z}{2}\,\coth\left(\frac{z}{2}\right)\tag{2}$$ hence, by exploiting $\int_{0}^{+\infty}z^{2k-1}e^{-z^2}\,dz = \frac{1}{2}(k-1)!$ we have: $$ \sum_{k\geq 1}\frac{(k-1)!}{(2k)!}\,B_{2k}=\int_{0}^{+\infty}e^{-z^2}\left(\coth\frac{z}{2}-\frac{2}{z}\right)\,dz=0.082006\ldots\tag{3}$$