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How to get closed form for the sum $\displaystyle{\sum\limits_{k = 1}^\infty {\frac{{{p^k}}} {{\left( {2k} \right)!!}}\frac{{{d^k}}} {{d{s^k}}}{{\left. {\frac{1} {{\Gamma \left( s \right)}}} \right|}_{s = \frac{3} {2}}}} }$ for $0 < p \leqslant 1$? For $p=1$, I have managed to obtain (numerically and by guesswork) $\displaystyle\sum\limits_{k = 1}^\infty {\frac{1} {{\left( {2k} \right)!!}}\frac{{{d^k}}} {{d{s^k}}}{{\left. {\frac{1} {{\Gamma \left( s \right)}}} \right|}_{s = \frac{3} {2}}}} = 1 - \frac{1} {{\Gamma \left( {\frac{3} {2}} \right)}}$ which would imply $\displaystyle\sum\limits_{k = 0}^\infty {\frac{1} {{\left( {2k} \right)!!}}\frac{{{d^k}}} {{d{s^k}}}{{\left. {\frac{1} {{\Gamma \left( s \right)}}} \right|}_{s = \frac{3} {2}}}} = 1$. I have no idea where to start.

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    $\begingroup$ Since $(2k)!!=2^k k!$ the series is the Taylor series expansion of $1/\Gamma((3+p)/2)$ at $p=0$ $\endgroup$ – Heike Jan 13 '12 at 9:35
  • $\begingroup$ So the answer would be $\frac{1} {{\Gamma \left( {\frac{{3 + p}} {2}} \right)}} - \frac{1} {{\Gamma \left( {\frac{3} {2}} \right)}}$. Thank you, very clever, I'm tired so didn't see it. If you want to make it into the answer, I will accept it and upvote it $\endgroup$ – Alen Jan 13 '12 at 9:58
  • $\begingroup$ I've posted my comment as an answer now. $\endgroup$ – Heike Jan 13 '12 at 12:20
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By writing $(2k)!!$ as $(2k)!!=2^k k!$ the series becomes $$ \sum_{k=1}^\infty \frac{1}{k!}\left(\frac{p}{2}\right)^k \left.\frac{d^k}{ds^k}\frac{1}{\Gamma(s)}\right|_{s=\frac{3}{2}} $$ which is by definition the Taylor series expansion of $$ \frac{1}{\Gamma\left(\frac{3+p}{2}\right)}-\frac{1}{\Gamma\left(\frac{3}{2}\right)} $$ at $p=0$.

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