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.
 A: $$  \begin{aligned}  \sum_{n\geq 0}\frac{(4n)!}{(4n+4)!}  & =\sum_{n\geq 0} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\& =\sum_{n\geq 0}\frac{1}{6}\mathrm{B}(4n+1, \,4) \\& = \frac{1}{6}\sum_{n\geq 0}\int_{0}^{1}x^{4n}(1-x)^3\; dx\\& = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \; dx  \\& = \frac{\ln 2}{4}-\frac{\pi}{24} \end{aligned}$$
A: Not nearly as impressive as the other answer, but elementary:
$$\begin{align}
\frac{(4n)!}{(4n+4)!} &= \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}\\
&= \left(\frac{1}{4n+1} - \frac{1}{4n+2}\right)\left(\frac{1}{4n+3} - \frac{1}{4n+4}\right)\\
&= \frac{1}{6} \frac{1}{4n+1} - \frac{1}{2}\frac{1}{4n+2} + \frac{1}{2}\frac{1}{4n+3} - \frac{1}{6}\frac{1}{4n+4}\\
&= \frac{1}{3}\left(\frac{1}{4n+1} - \frac{1}{4n+2} + \frac{1}{4n+3} - \frac{1}{4n+4}\right) - \frac{1}{6}\left(\frac{1}{4n+2} - \frac{1}{4n+4}\right) - \frac{1}{6} \left(\frac{1}{4n+1} - \frac{1}{4n+3}\right)
\end{align}$$
It is well-known that
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \log 2,$$
and the first parenthesis comprises four consecutive terms of that series, with no overlap, so from that we obtain $\frac{\log 2}{3}$. From the second parentheses, we can pull out a factor of $\frac12$ from both terms, then we obtain
$$\frac{1}{12} \left(\frac{1}{2n+1} - \frac{1}{2n+2}\right)$$
which comprises two consecutive terms of the $\log 2$ series, again without overlap, so together these two yield
$$\left(\frac{1}{3} - \frac{1}{12}\right)\log 2 = \frac{\log 2}{4}.$$
Another well-known series is Leibniz series
$$\frac{\pi}{4} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$$
and the last parenthesis comprises two consecutive terms of that, once again without overlap.
Since all parenthesised terms are dominated by $\frac{1}{n^2}$, we can split and rearrange to obtain
$$\begin{align}
\sum_{n = 0}^\infty \frac{(4n)!}{(4n+4)!} &= \frac13 \sum_{n = 0}^\infty \left(\frac{1}{4n+1} - \frac{1}{4n+2} + \frac{1}{4n+3} - \frac{1}{4n+4}\right)\\
&\quad -\frac{1}{12}\sum_{n=0}^\infty \left(\frac{1}{2n+1} - \frac{1}{2n+2}\right)\\
&\quad - \frac16 \sum_{n=0}^\infty \left(\frac{1}{4n+1} - \frac{1}{4n+3}\right)\\
&= \frac{\log 2}{3} - \frac{\log 2}{12} - \frac16\frac{\pi}{4} = \frac{\log 2}{4} - \frac{\pi}{24}.
\end{align}$$
A: $$\begin{align*}
\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}
&=\sum_{n=0}^\infty\frac{1}{(4n+1)\times(4n+2)\times(4n+3)\times(4n+4)}
\\&
\end{align*}$$
And then decompose in simple element !
Use a fixed $N$ and evaluate the sum going from $0$ to $N$ by putting the 4 sums to the same start value and the same end value. Then everything will disapear.
A: $$\begin{align*}
\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}
&=\sum_{n=0}^\infty\frac{1}{(4n+1)\times(4n+2)\times(4n+3)\times(4n+4)}
\\&=\frac{1}{8}\sum_{n=0}^\infty\frac{1}{32 n^4+80 n^3+70 n^2+25 n+3}
\end{align*}$$
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:
$\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}$, $\frac{1}{8}\sum_{n=0}^\infty\frac{1}{32 n^4+80 n^3+70 n^2+25 n+3}$
