Sum to infinity (2)

Relevant wikis .- fractional binomial theorem and Negative binomial theorem

The power series of $(1 - 2x)^{\frac{-1}{2}} = \frac{1}{\sqrt{1 - 2x}}$ with $|x| < 1/2$ is : $1 + x + \frac{3x^2}{2} + \frac{5x^3}{2} + \frac{35x^4}{8} + \frac{63x^5}{8} + o(x^5) = \displaystyle \sum_{k = 0}^{\infty} {-1/2 \choose k} (-2x)^k =$ $=1 + (-1/2)\cdot(-2x) + \frac{(-1/2)\cdot (-3/2)}{2!} (-2x)^2 + \frac{(-1/2)\cdot(-3/2)\cdot(-5/2)}{3!}(-2x)^3 +....$ Now substituing $x = 1/3$ we get $1 + \frac{1}{3} + \frac{1\cdot 3}{3\cdot 6} + \frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9} + ... =$ $(1 - 2\cdot \frac{1}{3})^{\frac{-1}{2}} = \frac{1}{\sqrt{1 - 2\cdot\frac{1}{3}}} = \sqrt{3}$

Details.-

$3 = 3^1 \cdot 1!, \quad \\ 3 \cdot 6 = 3^2 \cdot 2!, \quad \\3 \cdot 6 \cdot 9 = 3^3 \cdot 3!...$

Note.- This problem is exactly equal to this

The general term is can be simplified as

$\dbinom{2n}{n}(x)^n$ ,where $x=1/6$ ,thus the required sum is

$\sum_0^\infty \dbinom{2n}{n}(x)^n$

now it is interesting to note that,

$\begin{aligned} {-1/2\choose n}&=&{(-1/2)(-1/2-1)(-1/2-2)\cdots(-1/2-(n-1))\over n!}\cr &=&{(-1)^n\over 2^n} {(1)(3)(5)\cdots(2n-1)\over n!}\cr &=&{(-1)^n\over 2^n} {(1)(3)(5)\cdots(2n-1)\over n!}\cdot{2^n n!\over 2^n n!}\cr &=&{(-1)^n\over 4^n} {2n\choose n}. \end{aligned}$

thus the binomial expansion of $(1-4x)^{\tfrac{-1}{2}}=\sum_0^\infty \dbinom{2n}{n}(x)^n$

thus plugging in $x=1/6$ we get,

$\sum_0^\infty \dbinom{2n}{n}(\cfrac{1}{6})^n=(1-4(\cfrac{1}{6}))^{\tfrac{-1}{2}}=\boxed{\sqrt{3}}$