TITLE: Ways to find $\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$
QUESTION [9 upvotes]: $$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$
is equal to?

My approach:
We can see that the $n^{th}$ term is \begin{align}a_n&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\\&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\color{red}{[(2n+2)-(2n+1)}]\\&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)}-\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)\cdot(2n+1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\\
\end{align}
From here I just have a telescopic series to solve, which gave me $$\sum_{n=1}^{\infty}a_n=0.5$$
Another approach : note : $$\frac{(2n)!}{2^nn!}=(2n-1)!!$$
Which gives $$a_n=\frac{1}{2}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right)$$
So basically I need to compute
$$\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right) \tag{*}$$
I'm not able to determine the binomial expression of $(*)$ (if it exists) or else you can just provide me the value of the sum
Any hints will be appreciated, and you can provide different approaches to the problem too

REPLY [5 votes]: We can write the series as
\begin{align*}
\color{blue}{\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right)}
&=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{2n}{n}\left(\frac{1}{4}\right)^n\tag{1}\\
&=\frac{1}{2}\sum_{n=1}^{\infty}C_n\left(\frac{1}{4}\right)^n\tag{2}\\
&=\frac{1}{2}\left(\left.\frac{1-\sqrt{1-4x}}{2x}\right|_{x=\frac{1}{4}}\right)-\frac{1}{2}\tag{3}\\
&=\frac{1}{2}\cdot 2-\frac{1}{2}\tag{4}\\
&\,\,\color{blue}{=\frac{1}{2}}
\end{align*}

Comment:

In (1) we write the coefficient using binomial coefficients.

In (2) we note that $C_n=\frac{1}{n+1}\binom{2n}{n}$ are the ubiquituous Catalan numbers.

In (3) we use the generating function of the Catalan numbers evaluated at $x=\frac{1}{4}$. Since the series expansion of the generating function starts with $n=0$ we compensate it by subtracting $\frac{1}{2}$.

in (4) we evaluate the series at $x=\frac{1}{4}$ and simplify in the last step.