summation - Find the sum of series $sum_{n=0}^inftyfrac{(4n)!}{(4n+4)!}$

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.

Answer

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}$$