Is the infinite sum $\sum_{s=2}^\infty \frac{\zeta(s)}{s!}$ known? If so, what is its value? I recently ran into this infinite sum:
$$\sum_{s=2}^\infty \frac{\zeta(s)}{s!}$$
and have tried to solve it to no avail. Any references, solutions, or general advice would be greatly appreciated.
 A: This is quite a roundabout way of linking the series to the form provided in user17762's comment, but still I'll post it.
$$\zeta(s)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x-1}dx$$
Interchanging summation and integration we have:
$$S=\sum_{s=2}^\infty \frac{\zeta(s)}{s!}=\int_0^\infty \frac{dx}{e^x-1} \sum_{s=2}^\infty \frac{x^{s-1}}{s!(s-1)!}$$
Now the series inside can be easily seen as:
$$\sum_{s=2}^\infty \frac{x^{s-1}}{s!(s-1)!}=\sum_{s=2}^\infty \frac{sx^{s-1}}{s!^2}=\frac{d}{dx} \left( I_0 (2\sqrt{x})-1-x \right)=\frac{1}{\sqrt{x}}  I_1 (2\sqrt{x})-1$$
Now we write:

$$S=\int_0^\infty \frac{dx}{e^x-1}  \left( \frac{1}{\sqrt{x}}  I_1 (2\sqrt{x})-1\right)$$

This integral is present in the MathWorld refernece from Raymond Manzoni's comment.
Let's expand:
$$\frac{1}{e^x-1} =\sum_{k=0}^\infty e^{-(k+1)x}$$
So:
$$S=\sum_{k=0}^\infty  \int_0^\infty \left( \frac{1}{\sqrt{x}}  I_1 (2\sqrt{x})-1\right) e^{-(k+1)x}  dx=$$
$$=\sum_{k=0}^\infty \left( \int_0^\infty \frac{1}{\sqrt{x}}  I_1 (2\sqrt{x})e^{-(k+1)x}  dx-\frac{1}{k+1} \right)$$
It can be seen using (for example) the integral definition of the Bessel function that:
$$\int_0^\infty \frac{1}{\sqrt{x}}  I_1 (2\sqrt{x})e^{-(k+1)x}  dx=2 \int_0^\infty I_1 (2y)e^{-(k+1)y^2}  dy=e^{1/(k+1)}-1$$
Finally we obtain:

$$S=\sum_{k=0}^\infty \left( e^{1/(k+1)}-1-\frac{1}{k+1} \right)$$

Just as user17762 derived from the series definition for the Zeta function.
