<s>[INST] How to show $\sum_{n=0}^\infty (-1)^n\, \frac{\Gamma \left(\frac{n}{2}+1\right)}{n! \,\Gamma \left(2-\frac{n}{2}\right)}=\frac{3-\sqrt{5}}{2}$ 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?
 [/INST] 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*}
</s>