On $\sum_{k=1}^\infty1/(k!k^s)$.

Question

Has the following $\zeta$-like function been studied before?

$$f(z;s)=\sum_{k=1}^\infty\frac{z^k}{k!k^s}.$$

I believe this is an entire function since using the ratio test,

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{\frac{1}{(n+1)!(n+1)^s}}{\frac{1}{n!n^s}}=\lim_{n\to\infty}\frac{n!n^s}{(n+1)!(n+1)^s}=\lim_{n\to\infty}\frac{1}{n+1}\left(1+\frac{1}{n}\right)^{-s}=0,$$

so $R=\infty$.

The polylogarithm,

$$\text{Li}_s(z)=\sum_{k=1}^\infty\frac{z^k}{k^s},$$

"comes close," but does anyone know anything about $f(z;s)$, or similar function, in the literature?

@reuns comment reminded me of Ramanujan's Master Theorem. If we consider the slightly different function,

$$f_1(x;s)=\sum_{k=0}^\infty\frac{1}{k!(k+1)^s}(-x)^k,$$

then applying Ramanujan's Master Theorem with $\phi(k)=1/(1+k)^s$,

$$\int_0^\infty x^{s-1}f_1(x;s)dx=\frac{\Gamma(s)}{(1-s)^s}.$$

Applying the inverse Mellin transform then gives

$$f_1(x;s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{-s}\frac{\Gamma(s)}{(1-s)^s}ds,$$

assuming everything converges.

Answer 1

$$f(z;s)=\sum_{k=1}^\infty\frac{z^k}{k!k^s}.$$ $$f(z;0)=e^z-1$$ $$f(z;1)=Ei(z)-\ln(z)-\gamma$$ $$f(z;2)=z \:_3F_3(1,1,1;2,2,2;z)$$ In case of $s=n=$integer the closed form is a generalized hypergeometric function : $$f(z;n)=z \:_{n+1}F_{n+1}(1,...1;2,...,2;z)$$ $$f(z;-1)= ze^z$$ $$f(z;-2)= z(z+1)e^z$$ $$f(z;-3)= z(z^2+3z+1)e^z$$ In case of $-s=n=$integer the closed form is on the form of the next equation where $P(z)$ is a polynomial of degree $n-1$. $$f(z;-n)= zP(z)e^z$$ In the general case ($s$ real) it is doubtful that a closed form exists.