Evaluation of $\sum_{n=1}^\infty \frac{1}{\Gamma (n+s)}$ I want to try and evaluate this interesting sum:
$$\sum_{n=1}^\infty \frac{1}{\Gamma (n+s)}$$
where $0 \le s < 1$
WolframAlpha evaluates this sum to be 
$$\sum_{n=1}^\infty \frac{1}{\Gamma (n+s)} = e\left(1-\frac{\Gamma(s, 1)}{\Gamma(s)}\right)$$
Some notable cases of this sum would be when $s=0$ (producing the Taylor's series for $e$) and when $s=\frac{1}{2}$:
$$\sum_{n=1}^\infty \frac{1}{\Gamma (n+\frac{1}{2})} = e \operatorname {erf}(1)$$
I would be very interested to know the steps of how one would evaluate this interesting sum.
 A: I guess one starts by considering a more general sum:
$$E_{\alpha,\beta}(z) = \sum_{n=0}^\infty \frac{z^n}{\Gamma(\alpha n+\beta)}$$
which is known as a Mittag-Leffler function. For the special case of $\alpha=1$, the function satisfies a differential equation:
$$
    z \frac{d}{d z} E_{1,s}(z) + (s-1) E_{1,s}(z) = \sum_{n=0}^\infty \frac{(n+s-1)z^n}{\Gamma(n+s)} =z  \sum_{n=0}^\infty \frac{z^{n-1}}{\Gamma(n-1+s)} = \frac{1}{\Gamma(s-1)} + z E_{1,s}
$$
This is an inhomogeneous equation of the first order
$$
     z y^\prime(z) + (s-1-z) y(z) = \frac{1}{\Gamma(s-1)} 
$$
$$
     z \frac{\mathrm{d}}{\mathrm{d} z} \left( z^{s-1} \mathrm{e}^{-z} y(z) \right) = \frac{1}{\Gamma(s-1)} z^{s-1} \mathrm{e}^{-z}
$$
Hence
$$
    y(z) = \frac{1}{z^{s-1} \mathrm{e}^{-z}} \left( C - \frac{1}{\Gamma(s-1)} \int_z^\infty t^{s-2} \mathrm{e}^{-t} \mathrm{d} t \right) 
$$
The integral on the right hand-side is known as incomplete Gamma function.
Incidentally, the original series is also a hypergeometric series, meaning that $E_{1,s}(z)$ represents a hypergeometric function. Indeed:
$$
   E_{1,s}(z) =  \frac{1}{\Gamma(s)}{}_1F_1\left(1; s; z\right) = \frac{1}{\Gamma(s)} \sum_{n=0}^\infty \frac{(1)_n}{(s)_n} z^n = \sum_{n=0}^\infty \frac{z^n}{\Gamma(n+s)}
$$
where $(s)_n  = \frac{\Gamma(n+s)}{\Gamma(s)}$ was used.
A: If you're willing to start with the expansion of the lower incomplete gamma function discussed here:
$$\gamma(s, x) = x^s \, \Gamma(s) \, e^{-x}\sum_{n=1}^\infty\frac{x^n}{\Gamma(s+n)}
$$
Then:
$$\begin{align*}
\Gamma(s) &= \Gamma(s,1) + \gamma(s,1)
\\
e &= \frac{e\Gamma(s,1)}{\Gamma(s)} + \sum_{k=1}^\infty \frac{1}{\Gamma(s+n)}
\end{align*}
$$
A: We find an integral expression for the sum (the $u$ integral below) without appealing to the properties of special functions.
We have 
$$\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{\Gamma(n+s)}
&=& \frac{1}{\Gamma(s+1)} 
\underbrace{\left(1+\frac{1}{s+1}+\frac{1}{(s+1)(s+2)} + \ldots\right)}_{f(s)}.
\end{eqnarray*}$$
The series $f(s)$ is a simple example of an inverse factorial series. 
Such series were studied even in the 18th century by Nicole and Stirling and are dealt with, for example, in Whittaker and Watson's A Course of Modern Analysis.
One way to develop such a series is by successively integrating by parts the right hand side of
$$f(s) = \int_0^1 d\xi\, s(1-\xi)^{s-1} F(\xi),$$
where $F(\xi)$ is some analytic function of $\xi$ and 
$\int_0^1$ is shorthand for $\lim_{\epsilon\to 0^+}\int_0^{1-\epsilon}$.
One finds 
$$\begin{eqnarray*}
f(s) &=& F(0) + \frac{F'(0)}{s+1} + \frac{F''(0)}{(s+1)(s+2)} +\ldots.
\end{eqnarray*}$$
For details on the restrictions on $F(\xi)$, see Whittaker and Watson's 4th edition, $\S 7.82$.
For this problem we have $F^{(n)}(0) = 1$, so $F(\xi) = e^\xi$. 
Then 
$$\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{\Gamma(n+s)} &=& \frac{f(s)}{\Gamma(s+1)} \\ 
&=& \frac{1}{\Gamma(s+1)} \int_0^1 d\xi\, s(1-\xi)^{s-1} e^\xi \\
&=&  \frac{e}{\Gamma(s)} \int_0^1 du\, u^{s-1} e^{-u}
    \hspace{10ex}(\textrm{let }u=1-\xi) \\
&=& \frac{e}{\Gamma(s)} \gamma(s,1),
\end{eqnarray*}$$
where $\gamma(s,x)$ is the lower incomplete gamma function.
Note that $\gamma(s,x) = \Gamma(s) - \Gamma(s,x)$, where $\Gamma(s,x)$ is the upper incomplete gamma function.
Therefore, 
$$\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{\Gamma(n+s)} 
&=& e\left(1-\frac{\Gamma(s,1)}{\Gamma(s)}\right),
\end{eqnarray*}$$
as claimed. 
Thanks for the interesting question! 
