Prove that $1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$ 
Prove that $$1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$$

$\bf{My\; Try::}$ Using Binomial expansion of  $$(1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+.......$$
So we get $$nx=\frac{1}{3}$$ and $$\frac{nx(nx+x)}{2}=\frac{1}{3}\cdot \frac{3}{6}$$
We get $$\frac{1}{3}\left(\frac{1}{3}+x\right)=\frac{1}{3}\Rightarrow x=\frac{2}{3}$$
So we get $$n=\frac{1}{2}$$
So our series sum is $$(1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$
Although I know that this is the simplest proof, can we solve it any other way someting like defining $a_{n}$ and then use Telescopic sum.
Thanks.
 A: Here is a variation to obtain $\sqrt{3}$ based upon the generating function of the Central binomial coefficients
\begin{align*}
  \sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad\qquad |z|<\frac{1}{4}
  \end{align*}

We obtain
\begin{align*}
  1&+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+\cdots\\
  &=1+\frac{1!!}{3^11!}+\frac{3!!}{3^22!}+\frac{5!!}{3^33!}+\cdots\tag{1}\\
  &=1+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n!}\frac{1}{3^n}\\
  &=1+\sum_{n=1}^{\infty}\frac{(2n)!}{n!(2n)!!}\frac{1}{3^n}\tag{2}\\
  &=1+\sum_{n=1}^{\infty}\frac{(2n)!}{n!n!}\frac{1}{6^n}\tag{3}\\
  &=\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{6^n}\\
  &=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=\frac{1}{6}}\\
  &=\frac{1}{\sqrt{1-\frac{2}{3}}}\\
  &=\sqrt{3}
  \end{align*}

Comment:


*

*In (1) we use double factorial for odd values $$(2n-1)!!=(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1$$

*In (2) we use the identity
\begin{align*}
  (2n)!=(2n)!!(2n-1)!!
  \end{align*}

*In (3) we use the identity
\begin{align*}
  (2n)!!=2^nn!
  \end{align*}

