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

$$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$ 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 $$\frac{1}{2}\sum_{r=1}^{\infty}\frac{(2r)!}{4^{r}r!(r+1)!}=\frac{1}{2}\sum_{r=0}^{\infty}\frac{(2r)!}{4^{r}r!(r+1)!}-\frac{1}{2}(1)=1-\frac{1}{2}=\frac{1}{2}$$

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.

I tried to rewrite the sum:

$$\sum _{n=1}^k \frac{\left(\frac{1}{4}\right)^n (2 n)!}{n! (n+1)!}$$

computationally using the Mathematica Code:

Sum[Factorial[2*n]*(1/4)^n/(Factorial[n]*Factorial[n + 1]), {n, 1, k}]

As a result I got:

$$\frac{2^{-2 k-1} \left(-k (2 (k+1))!-2 (2 (k+1))!+2^{2 k+1} (k+1)! (k+2)!\right)}{(k+1)! (k+2)!}\\=1-\frac{2^{-2 k-1} (k+2) (2 (k+1))!}{(k+1)! (k+2)!}=1-\frac{2 \Gamma \left(k+\frac{3}{2}\right)}{\sqrt{\pi } \Gamma (k+2)}$$

where $\Gamma$ is Euler Gamma Function. By setting $k=\infty$, the complete sum becomes 1.