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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: $$S=\sum_{k=1}^\infty e^{1/k}-\left(1+\frac1k\right)$$

Henry Lee
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    $\sum_{k=1}^\infty \frac1{k^n}$ is not geometric but $\sum_{n=1}^\infty \frac1{k^n}$ is. – Occam Dec 02 '20 at 17:08
  • You can change the order of summation btw. This follows because $S$ is convergent and the convergence is absolute. – dude Dec 02 '20 at 17:16
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    Expanding $\zeta(n)$ into its series representation and switching the order of summation before summing, you find that your sum is equal to $\sum_{k\ge 1} e^{1/k}-1-1/k.$ It converges because the summand is $\frac{1}{2k^2} +\mathcal{O}\left(\frac{1}{k^3}\right)$ as $k\to\infty$, but I am not sure if there is a way to evaluate it in closed-form. – Diffusion Dec 02 '20 at 17:23

3 Answers3

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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 ...

Occam
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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.

Max Lonysa Muller
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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.

Infinity_hunter
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