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.

  • 2
    $\begingroup$ 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

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 $p=0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.