# An infinite series

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• July 1st 2009, 10:56 PM
simplependulum
An infinite series
We know that :

$\sum_{k=0}^{\infty} ( \frac{x}{4} )^k \binom{2k}{k} = \frac{1}{ \sqrt{ 1 - x}} , |x| < 1$

I'd like to ask what about

$\sum_{k=0}^{\infty} ( \frac{x}{16} )^k \binom{4k}{2k} , |x| < 1 ~~$ ?
• July 4th 2009, 07:10 AM
Jester
Quote:

Originally Posted by simplependulum
We know that :

$\sum_{k=0}^{\infty} ( \frac{x}{4} )^k \binom{2k}{k} = \frac{1}{ \sqrt{ 1 - x}} , |x| < 1$

I'd like to ask what about

$\sum_{k=0}^{\infty} ( \frac{x}{16} )^k \binom{4k}{2k} , |x| < 1 ~~$ ?

If

$\frac{1}{ \sqrt{ 1 - x}} = \sum_{k=0}^{\infty} ( \frac{x}{4} )^k \binom{2k}{k} = 1 + \binom{2}{1} \frac{x}{4} + \binom{4}{2} \left(\frac{x}{4}\right)^2 + \binom{6}{3} \left(\frac{x}{4}\right)^3 + \binom{8}{4} \left(\frac{x}{4}\right)^4 \cdots$

then

$\frac{1}{ \sqrt{ 1 + x}} = 1 - \binom{2}{1} \frac{x}{4} + \binom{4}{2} \left(\frac{x}{4}\right)^2 - \binom{6}{3} \left(\frac{x}{4}\right)^3 + \binom{8}{4} \left(\frac{x}{4}\right)^4 \cdots$

then

$\frac{1}{2}\left( \frac{1}{ \sqrt{ 1 - x}}+ \frac{1}{\sqrt{ 1 + x}}\right) = 1 + \binom{4}{2} \left(\frac{x}{4}\right)^2 + \binom{8}{4} \left(\frac{x}{4}\right)^4 \cdots$

then replace $x$ with $\sqrt{x}$.