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.

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:
$$\begin{align*}
\Gamma(s) &= \Gamma(s,1) + \gamma(s,1)
\\
e &= \frac{e\Gamma(s,1)}{\Gamma(s)} + \sum_{k=1}^\infty \frac{1}{\Gamma(s+n)}
\end{align*}
$$
