Closed form for $\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}$

Question

What do you think about the closed form I found for this series?

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

I found it this way:

Take the Ramanujan's summation $C(0)$ for divergent series:

$$C(0) = -\frac{1}{2}f(0) - \sum_{k=1}^\infty \frac{ B_{2k}}{(2k)!}f^{(2k-1)}(0)$$

Let $f(x)=e^x$ so that $f^{(2k-1)}(0)=f(0)=1$.

Now we have:

$$C(0) = -\frac{1}{2} - \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}$$

Lets now assign some value to $C(0)$, by manipulating $\sum_{k=1}^\infty e^k$ one arrives to:

$$S = \sum_{k=1}^\infty e^k = e^1+e^2+e^3+...$$ $$\frac{1}{e}S=1+S$$ $$S=\frac{e}{1-e}$$

Now we have:

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

Answer 1

It seems that the identity follows immediately from the definition of Bernoulli number. It si known that for $|x|<2\pi$, $$\frac{x}{e^x-1}+\frac{x}{2}-1 = \sum_{k=1}^\infty \frac{B_{2k}x^{2k}}{(2k)!}$$ so by taking $x=1$ we get the result (different from OP's) $$\frac{3-e}{2(e-1)}=\frac{1}{e-1}+\frac{1}{2}-1 = \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}.$$ Am I missing something?

P.S.By the way the $$\frac{e+1}{2\left(e-1\right)}=\sum_{k=0}^\infty \frac{B_{2k}x^{2k}}{(2k)!}.$$

Answer 2

What do I think? That you should take care when using Ramanujan Summation this way.

Note that: $$\frac{e+1}{2\left(e-1\right)} =\color{red}{1}.08197670687\ldots\neq \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} = \color{red}{0}.08197670687\ldots$$

Without using numerics we can already see that $$\frac{e+1}{2\left(e-1\right)} \gt \frac{e+1}{2\left(e+1\right)} = 1/2$$ While we have $$\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!} < \sum_{k=1}^2\frac{B_{2k}}{(2k)!} = \frac{59}{720}<1/2$$ Proving that the two are not equal.

Answer 3

According to this page, a Taylor series is $$\coth x=\dfrac{1}{x}+\sum\limits_{n=1}^\infty\dfrac{4^nB_{2n}}{(2n)!}x^{2n-1}\qquad \text{for} \qquad 0<|x|< \pi$$ where $B_n$ is the $n$th Bernoulli number.

We can rewrite it as $$\coth x=\dfrac{1}{x}+\dfrac{1}{x}\sum\limits_{n=1}^\infty\dfrac{B_{2n}}{(2n)!}(2x)^{2n}$$ So, making $x=\frac 12$ we have $$\coth\left(\frac 12\right)=\frac{e^{1/2}+e^{-1/2}}{e^{1/2}-e^{-1/2}}=\frac{e+1}{e-1}=2+2\sum\limits_{n=1}^\infty\dfrac{B_{2n}}{(2n)!}$$ and then $$\sum\limits_{n=1}^\infty\dfrac{B_{2n}}{(2n)!}=\frac{3-e}{2 (e-1)}$$