Series involving gamma function

Question

In the study of gamma incomplete function

does any know how to solve

$\sum_{k=0}^{\infty} \frac{1}{\Gamma(k+s+1)}$

Answer 1

Note that $$ \eqalign{ & \sum\nolimits_{k = 0}^{\;\infty } {{1 \over {\Gamma \left( {s + 1 + k} \right)}}} = \sum\nolimits_{k = s}^{\;\infty } {{1 \over {\Gamma \left( {1 + k} \right)}}} = \cr & = \sum\nolimits_{k = 0}^{\;\infty } {{1 \over {k!}}} - \sum\nolimits_{k = 0}^{\;s} {{1 \over {k!}}} = \cr & = e - e\left( {{1 \over e}\sum\nolimits_{k = 0}^{\;s} {{1 \over {k!}}} } \right) = \cr & = e\left( {1 - Q\left( {s,1} \right)} \right) = e\left( {1 - {{\Gamma \left( {s,1} \right)} \over {\Gamma \left( s \right)}}} \right) \cr} $$ where:
- the summation is intended in the definition as Antidifference;
- $Q(s,z)$ is the Regularized Upper Incomplete Gamma, which is known to represent the ratio of the partial to complete expansion of $exp(z)$ (it follows from its functional definition).

In another way, through the definition of Rising Factorial $$ \eqalign{ & \sum\nolimits_{k = 0}^{\;\infty } {{1 \over {\Gamma \left( {s + 1 + k} \right)}}} = {1 \over {\Gamma \left( {s + 1} \right)}}\sum\nolimits_{k = 0}^{\;\infty } {{1 \over {{{\Gamma \left( {s + k + 1} \right)} \over {\Gamma \left( {s + 1} \right)}}}}} = \cr & = {1 \over {s!}}\sum\nolimits_{k = 0}^{\;\infty } {{1 \over {\left( {s + 1} \right)^{\,\overline {\,k\,} } }}} = {1 \over {s!}}\sum\nolimits_{k = 0}^{\;\infty } {{{1^{\,\overline {\,k\,} } } \over {\left( {s + 1} \right)^{\,\overline {\,k\,} } }}} {1 \over {k!}} = \cr & = {1 \over {s!}}{}_1F_{\,1} \left( {\left. {\matrix{ 1 \cr {s + 1} \cr } \;} \right|\;1} \right) \cr} $$ we get a Confluent Hypergeometric function computed at $x=1$, which confirms the above (re. e.g. to Wolfram Function site)