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

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 $20\text{%}$
