Sum of a power series with a parameter Does anybody know the sum

$$ \sum_{n=0}^{\infty}\frac{x^{n}}{(n+a)n!}=f(x,a)$$

here $a$ is a number $ a&gt;0 $. A hint please ? :D

If $ a=1$ I believe $ f(x,1)=(e^{x}-1)/x $
For $a\gt0$,
$$
\begin{align}
af(x,a)
&amp;=a\sum_{n=0}^\infty\frac{x^n}{(n+a)n!}\\
&amp;=1+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+a}\right)\frac{x^n}{(n-1)!}\\
&amp;=e^x-x\sum_{n=0}^\infty\frac{x^n}{(n+a+1)n!}\\
&amp;=e^x-xf(x,a+1)\tag{1}
\end{align}
$$
Reversing $(1)$ yields
$$
f(x,a)=\frac{e^x-(a-1)f(x,a-1)}{x}\tag{2}
$$
Evaluating $(1)$ directly gives
$$
\begin{align}
f(x,1)
&amp;=\sum_{n=0}^\infty\frac{x^n}{(n+1)n!}\\
&amp;=\frac1x\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}\\
&amp;=\frac{e^x-1}{x}\tag{3}
\end{align}
$$
$f(x,a)$ can be computed for higher $a\in\mathbb{Z}$ using recursion and $(2)$.
I think you should rewrite your function in a better way...
$$
f(x,a) = \sum \frac{x^n}{(n+a)n!} = \sum \frac{x^n}{n \cdot n!} + \sum \frac{x^n}{a \cdot n!} = S + \frac{1}{a}\sum \frac{x^n}{n!} = S + \frac{e^x}{a}.
$$
If you differentate $S$, you'll find that $S' = \sum \frac{n \cdot x^{n-1}}{n \cdot n!} = \sum \frac{x^{n-1}}{n!}$, so $xS' = \sum \frac{x^n}{n!} = e^x$.
Now ... since $xS' = e^x, S = \int e^x / x = Ei(x)$.
So $f(x,a) = \sum \frac{x^n}{(n+a)n!} = e^x/a + Ei(x)$.