# Ways to find $\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$

binomial theoremfactorialsequences-and-series

$$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$
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

If you look at the Binomial expansion of

$$(1-x)^{-\frac{1}{2}}$$ you get :-

$$\sum_{r=0}^{\infty}\frac{\binom{2r}{r}x^{r}}{4^{r}}$$

So $$\int_{0}^{1}(1-x)^{-\frac{1}{2}}dx=\sum_{r=0}^{\infty}\frac{\binom{2r}{r}}{4^{r}(r+1)}$$

So you get $$\frac{1}{2}\sum_{r=0}^{\infty}\frac{\binom{2r}{r}}{4^{r}(r+1)}=\frac{1}{2}\sum_{r=0}^{\infty}\frac{(2r)!}{4^{r}r!(r+1)!}=\frac{1}{2}\int_{0}^{1}(1-x)^{-\frac{1}{2}}dx=1$$

So $$\frac{1}{2}\sum_{r=1}^{\infty}\frac{(2r)!}{4^{r}r!(r+1)!}=\frac{1}{2}\sum_{r=0}^{\infty}\frac{(2r)!}{4^{r}r!(r+1)!}-\frac{1}{2}(1)=1-\frac{1}{2}=\frac{1}{2}$$