Find the sum of series $\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}$ I wanted to know how can I start to find the sum of the series:

$$\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}=\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}\cdots$$

I am having no clue.

Thanks.
$$  \begin{aligned}  \sum_{n\geq 0}\frac{(4n)!}{(4n+4)!}  &amp; =\sum_{n\geq 0} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\&amp; =\sum_{n\geq 0}\frac{1}{6}\mathrm{B}(4n+1, \,4) \\&amp; = \frac{1}{6}\sum_{n\geq 0}\int_{0}^{1}x^{4n}(1-x)^3\; dx\\&amp; = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \; dx  \\&amp; = \frac{\ln 2}{4}-\frac{\pi}{24} \end{aligned}$$
$$\begin{align*}
\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}
&amp;=\sum_{n=0}^\infty\frac{1}{(4n+1)\times(4n+2)\times(4n+3)\times(4n+4)}
\\&amp;=\frac{1}{8}\sum_{n=0}^\infty\frac{1}{32 n^4+80 n^3+70 n^2+25 n+3}
\end{align*}$$

Now that we have replaced the nasty factorials with a nice, clean, rational function, we can use a more general procedure from here.

Also, if all you need to do is prove convergence, the comparison test is the simplest conclusive test.

If you want some additional stuff, here are the Wolfram Alpha queries for the equations:
$\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}$, $\frac{1}{8}\sum_{n=0}^\infty\frac{1}{32 n^4+80 n^3+70 n^2+25 n+3}$