finding a value for $\sum_{n=2}^\infty\frac{\zeta(n)}{n!}$ 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:
$$S=\sum_{k=1}^\infty e^{1/k}-\left(1+\frac1k\right)$$
 A: It's a positive-term series, so rearrangements are fine (Baby Rudin, Ch.8, #3).
However, you won't get the $\frac1{n!}$ out of a series in $n$, so you'll have $\sum_{n=2}^\infty \frac{(1/k)^n}{n!}$. Not a geometric series, but still one that you should recognize ...
A: 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.
A: Well, here is an approximate value of $S$.
Observe that as $n$ gets bigger $\zeta(n)$ tends to $1$ moreover $\zeta(n)$ is decreasing sequence.  So for large $n$ we can consider $\zeta(n)$ to be approximately equal to $1$.
So
$$
S \approx \sum_{n=2}^{N} \frac{\zeta(n)}{n!} + \sum_{n = N+1}^{\infty} \frac{1}{n!} \approx e - 2 + \sum_{n = 2}^{N} \frac{\zeta(n) - 1}{n!} 
$$
Actually $\zeta(n)$ quickly converge to $1$ as $n$ becomes bigger. For example $\zeta(6) = \pi^6/945 \approx  1.017... $ and  $\zeta(8) = \pi^8/9450 \approx $1.004...
So depending on the need of accuracy you need to choose $N$ and can calculate the value of $S$.
Bonus thing : We have an exact expression for zeta function at even positive integers given by
$$
\zeta(2n) = (-1)^{n-1}\frac{(2\pi)^{2n}B_{2n}}{2(2n)!}
$$
where $B_n$ are Bernoulli's number. I leave it to you use this if you need more exact expression even though we do not know have any closed expression at odd integers.
