# How to find sum of the given series?

## Jul 11, 2018

$\sqrt{\frac{5}{3}}$

#### Explanation:

From the binomial expansion of ${\left(1 - x\right)}^{- \frac{1}{2}}$ we get

${\left(1 - x\right)}^{- \frac{1}{2}} = 1 + \left(- \frac{1}{2}\right) \left(- x\right)$
qquadqquadqquadqquad quad +1/(2!)(-1/2)(-3/2)(-x)^2+...
qquadqquadqquadqquad =1+1/2 x+1/(2!)* 1/2*3/2x^2+...

we see that the coefficient of ${x}^{n} , \setminus \setminus n \ge 1$ is

1/(n!) *1/2*3/2*5/2*...*(2n-1)/2
=1/(n!) (1*3*5*...(2n-1))/2^n
= 1/(n!) (1*2*3*...*(2n-1)*(2n))/(2^n*2*4*6*...(2n))

= ((2n)!)/(2^{2n}(n!)^2)= 1/4^n((2n),(n))

Thus

${\left(1 - x\right)}^{- \frac{1}{2}} = 1 + {\sum}_{n = 1}^{\infty} {\left(\frac{x}{4}\right)}^{n} \left(\begin{matrix}2 n \\ n\end{matrix}\right) = {\sum}_{n = 0}^{\infty} {\left(\frac{x}{4}\right)}^{n} \left(\begin{matrix}2 n \\ n\end{matrix}\right)$

and so

${\sum}_{n = 1}^{\infty} {x}^{n} \left(\begin{matrix}2 n \\ n\end{matrix}\right) = {\left(1 - 4 x\right)}^{- \frac{1}{2}}$

So, our sum

${\sum}_{n = 1}^{\infty} \frac{1}{10} ^ n \left(\begin{matrix}2 n \\ n\end{matrix}\right) = {\left(1 - 4 \times \frac{1}{10}\right)}^{- \frac{1}{2}} = \sqrt{\frac{5}{3}}$