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The following sum was used in an unrelated answer on math.stackexchange.com.

$$\sum_{n=0}^\infty (-1)^n \frac{\Gamma \left(\frac{n}{2}+1\right)}{n! \Gamma \left(2-\frac{n}{2}\right)}=\frac{3-\sqrt{5}}{2}$$

How can you show this to be true?

  • Note: we must be careful. For even values of $n\ge4$, the summand must be interpreted as zero due to the pole of the gamma function. – FShrike Oct 26 '22 at 09:17
  • I suggest you expand the terms manually into double factorials. It’ll hopefully boil down to something convenient. – FShrike Oct 26 '22 at 09:19
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    What do you get after using $$ \Gamma \left( {\frac{1}{2} + m} \right) = \frac{{(2m)!}}{{4^m m!}}\sqrt \pi ,\quad \Gamma \left( {\frac{1}{2} - m} \right) = \frac{{( - 4)^m m!}}{{(2m)!}}\sqrt \pi ;? $$ – Gary Oct 26 '22 at 09:23
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    @Gary I am not sure how to use those facts. Would you be able to add an answer please? –  Oct 27 '22 at 13:51

2 Answers2

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We will use the facts that $$ \frac{1}{{\Gamma\! \left( {2 - \frac{n}{2}} \right)}} = \frac{{\Gamma \!\left( {\frac{n}{2} - 1} \right)}}{{\Gamma \!\left( {2 - \frac{n}{2}} \right)\Gamma\! \left( {\frac{n}{2} - 1} \right)}} = - \frac{1}{\pi }\sin \left( {\frac{{\pi n}}{2}} \right)\Gamma \!\left( {\frac{n}{2} - 1} \right) $$ for $n\geq 3$, $$ \Gamma\! \left( {m + \frac{1}{2}} \right) = \frac{{(2m)!}}{{4^m m!}}\sqrt \pi $$ for $m\geq 0$, and $$ \sqrt {1 + x} = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^{k + 1} }}{{4^k (2k - 1)}}} \binom{2k}{k}x^k $$ for $|x|<1$. Thus \begin{align*} \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{\Gamma\! \left( {\frac{n}{2} + 1} \right)}}{{n!\Gamma\! \left( {2 - \frac{n}{2}} \right)}}} & = \frac{1}{2} + \sum\limits_{n = 3}^\infty {( - 1)^n \frac{{\Gamma\! \left( {\frac{n}{2} + 1} \right)}}{{n!\Gamma\! \left( {2 - \frac{n}{2}} \right)}}} \\ & = \frac{1}{2} + \frac{1}{\pi }\sum\limits_{n = 3}^\infty {( - 1)^{n + 1} \sin \left( {\frac{{\pi n}}{2}} \right)\frac{{\Gamma\! \left( {\frac{n}{2} - 1} \right)\Gamma\! \left( {\frac{n}{2} + 1} \right)}}{{n!}}} \\ & = \frac{1}{2} + \frac{1}{\pi }\sum\limits_{k = 1}^\infty {( - 1)^k \frac{{\Gamma\! \left( {k - \frac{1}{2}} \right)\Gamma\! \left( {k + \frac{3}{2}} \right)}}{{(2k + 1)!}}} \\ & = \frac{1}{2} + \sum\limits_{k = 1}^\infty { \frac{( - 1)^k}{{(2k + 1)!}}\frac{1}{{16^k }}\frac{{(2k - 2)!}}{{(k - 1)!}}\frac{{(2k + 2)!}}{{(k + 1)!}}} \\ & = \frac{1}{2} + \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{16^k (2k - 1)}}} \binom{2k}{k} \\ & = \frac{3}{2} + \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{16^k (2k - 1)}}} \binom{2k}{k} \\ & = \frac{3}{2} - \sqrt {1 + \frac{1}{4}} = \frac{3}{2} - \frac{{\sqrt 5 }}{2}. \end{align*}

Gary
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  • This is great, thank you. If you make the sum more general e.g. $\sum_{n=0}^\infty (-1)^n \frac{\Gamma \left(kn+1\right)}{n! \Gamma \left((k-1)n + 2\right)}$ is there always going to be a nice closed form expression? –  Oct 27 '22 at 15:25
  • @graffe I am not sure, that will involve the Fox–Wright function. Did I answer your original question? – Gary Oct 31 '22 at 06:04
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$$\sum_{n=0}^\infty (-1)^n \frac{\Gamma \left(\frac{n}{2}+1\right)}{n!\, \Gamma \left(2-\frac{n}{2}\right)}x^n=\frac{1}{2} \left(x^2+2-x\sqrt{x^2+4} \right)$$

Claude Leibovici
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