I have been trying to find the sum:
$$S=\sum_{n=2}^\infty\frac{\zeta(n)}{n!}$$
And I can't find if anyone else has posted about this. Interestingly wolfram alpha claims it converges for $n=0,1$ as well which cannot be correct. I think maybe there is some way of solving this by changing the order of summation:
$$S=\sum_{n=2}^\infty\frac 1{n!}\sum_{k=1}^\infty\frac{1}{k^n}$$
but I am struggling to justify this move, also trying to find:
$$\sum_{n=2}^\infty\frac{1}{k^n}$$
seems to be a geometric series so can I just do it like so, or will the bounds of the summation have changed? Thanks
From the advice of answers I have:
$$\sum_{n=2}^\infty\frac{(1/k)^n}{n!}=\sum_{n=0}^\infty\frac{(1/k)^n}{n!}-\frac1k-1=e^{1/k}-\frac1k-1$$
so we now have:
$$S=\sum_{k=1}^\infty e^{1/k}-\left(1+\frac1k\right)$$
Best Answer
At the bottom of this page from Wolfram Mathworld on the Riemann zeta function, it is stated that the constant you define - which amounts to equation $(133)$ of the article and which I'll call $C_1$ here - does not have a closed form. Similar constants, like
\begin{align*} C_{2} &:= \sum_{n=1}^{\infty} \frac{\zeta(2n)}{n!} \\ & \approx 2.407446554790328514709486656223022725582266, \text{ and} \\ C_{3} &:= \sum_{n=1}^{\infty} \frac{\zeta(2n)}{(2n)!} \\ &\approx 0.869001991962908998811054805561395688892494\end{align*} which are defined in equations $(134)$ and $(137)$, don't have a closed form either.
In the following MSE question I ask for conjectured closed forms of such constants. As of September 2, 2022, no serious answers have been given yet.