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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.

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According to Maple, $\Gamma(s,1)$ can also be expressed, for example, as $Ei(1-s,1)$, or as $\Gamma \left( s \right) -{\frac {{\it LaguerreL} \left( -s,s,-1 \right) \pi }{\sin \left( \pi \,s \right) }} $ (if $s$ is not an integer). – Robert Israel Jul 2 '12 at 4:39
up vote 9 down vote accepted

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!

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Changed the title accordingly, thanks! – Argon Jul 2 '12 at 13:38
@Argon: Glad to help. Cheers! – user26872 Jul 2 '12 at 16:39

If you're willing to start with the expansion of the lower incomplete gamma function discussed here:

$$\gamma(s, x) = x^s \, \Gamma(s) \, e^{-x}\sum_{n=1}^\infty\frac{x^n}{\Gamma(s+n)} $$


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

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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.

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