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

Question

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?

Answer 1

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*}

Answer 2

$$\sum_{n=0}^\infty (-1)^n \frac{\Gamma \left(\frac{n}{2}+1\right)}{n!\, \Gamma \left(2-\frac{n}{2}\right)}x^n=\frac{1}{2} \left(x^2+2-x\sqrt{x^2+4} \right)$$

Answer 3

I would like to add a bit of detail to @ClaudeLeibovici's answer. Following @Gary's initial steps, $$F(x) :=\sum_{n=0}^\infty (-1)^n \frac{\Gamma\left(\tfrac{n}{2}+1\right)}{\Gamma\left(2-\tfrac{n}{2}\right)}\frac{x^n}{n!}= 1-x+\frac{1}{2}x^2+\sum_{n=3}^\infty (-1)^n \frac{\Gamma\left(\tfrac{n}{2}+1\right)}{\Gamma\left(2-\tfrac{n}{2}\right)}\frac{x^n}{n!}$$ the series over $n\ge3$ has vanishing even terms due to the pole of the $\Gamma$-function so, $$F(x) = 1-x+\frac{1}{2}x^2 -\sum_{n=1}^\infty\frac{\Gamma\left(n+\tfrac{3}{2}\right)}{\Gamma\left(\tfrac{3}{2}-n\right)}\frac{x^{2n+1}}{(2n+1)!}.$$ Now, by the duplication formula, $$\frac{1}{(2n+1)!} = \frac{\sqrt \pi }{ 2^{1+2n}n!\Gamma\left(n+\tfrac{3}{2}\right)}$$ for $n\ge 1$, so, $$\begin{align*}F(x)&=1-x+\frac{1}{2} x^2-\sqrt \pi\sum_{n=1}^\infty\frac{1}{2^{1+2n}\Gamma\left(\tfrac{3}{2}-n\right)}\frac{x^{2n+1}}{n!}\\ &=1-x+\frac{1}{2}x^2 - x\sum_{n=1}^\infty\binom{\tfrac{1}{2}}{n}\left(\frac{x}{2}\right)^{2n } \end{align*}$$ which by, $$\sqrt{1+t} =1+ \sum_{n=1}^\infty\binom{\tfrac{1}{2}}{n}t^n,\quad |t|<1$$ gives, $$F(x) = 1+\frac{1}{2}x^2-x\sqrt{1+\frac{x^2}{4}},\quad |x|<2$$ which is @ClaudeLeibovici's result.