Calculate (or estimate) $S(x)=\sum_{k=1}^\infty \frac{\zeta(kx)}{k!}$.

Question

Let $x\in\mathbb R$, $x>1$ and $$S(x)=\sum_{k=1}^\infty \frac{\zeta(kx)}{k!}$$ where $\zeta(x)$ is the Riemann zeta function. Calculate (or estimate) $S(x)$.

Answer 1

What one could do fairly easy (pure heuristics) is to obtain an expansion for large $x$. The Riemann Zeta function can be approximated in this limit by $\zeta(z)\sim 1+\frac{1}{2^z}$

Therefore our sum reads

$$ S(x)=\sum_{k=1}^{\infty}\frac{\zeta(kx)}{k!}\sim_{x\rightarrow\infty}\sum_{k=1}^{\infty}\frac{1}{k!}+\frac{1}{k! 2^{kx}}=e-2+e^{\frac{1}{2^x}}\sim_{x\rightarrow\infty}e-1+\frac{1}{2x} $$

which fits extremly well even for values as small as $x\approx 5$.

Because the Riemann Zeta function is monotonically decreasing for $z>1$ we might bound the series in question by simply estimating $\zeta(xk)<\zeta(x)$ to get

$$ (e-1)<S(x)<\zeta(x)(e-1) $$

which by the sandwich lemma yields the same big $x$ limit as above

I would be really surprised if an closed form for this series exists, but who knows...:)

Appendix:

By a very similar reasoning it is possible to get an approximation for $x\rightarrow 1_+$. It is well known, that in the vicinity of $z=1$ the Riemann Zeta function posseses an Laurent expansion of the form

$$ \zeta(z)\sim_{z\rightarrow 1_+}=\frac{1}{z-1}+\gamma $$

where $\gamma$ is the Euler-Marschoni constant. Therefore $S(x)$ is cleary dominated by the first term of the sum

$$ S(x)\sim_{x\rightarrow 1_+}\frac{1}{x-1}+\gamma+R(x) $$

We might observe that the terms of the remainder $R(x)$ are given asymptotically by $ \zeta(k)/(k!)$ which is like $\gamma$ of $\mathcal{O}(1)$ . Therefore

$$ S(x)\sim_{x\rightarrow 1_+}\frac{1}{x-1}+\gamma+C $$

turns out to be a very good approximation in this limit ($C=\sum_2^{\infty}\zeta(k)/k!)$. Even if $x=1.5$ we are only of by something like $20\text{%}$