# 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

• It seems hard to calculate the binomial. The telescope is the standard way to do it. Commented Dec 17, 2021 at 6:58
• One way is to use the fact $\frac{1}{\sqrt{1-z}} = \sum\limits_{n=0}^\infty \binom{2n}{n}\left(\frac{z}{4}\right)^n$ for $|z| < 1$ and $\frac{1}{n+1} = \int_0^1 z^n dz$, this allow you to rewrite the sum as an integral... Commented Dec 17, 2021 at 7:16

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.