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.

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.
