# [SOLVED] Power Series Proof

• Feb 26th 2006, 11:32 AM
jallison88
[SOLVED] Power Series Proof
How would one prove that the sum, from n=0 to infinite, of the alternating power series (-1)^n/[3^n*(2n+1)] equals Pi/[2(3)^.5]? I'm having some difficulty conceptualizing the process.
• Feb 26th 2006, 11:59 AM
ThePerfectHacker
Quote:

Originally Posted by jallison88
How would one prove that the sum, from n=0 to infinite, of the alternating power series (-1)^n/[3^n*(2n+1)] equals Pi/[2(3)^.5]? I'm having some difficulty conceptualizing the process.

Quote:

Originally Posted by jallison88
How would one prove that the sum, from n=0 to infinite, of the alternating power series (-1)^n/[3^n*(2n+1)] equals Pi/[2(3)^.5]? I'm having some difficulty conceptualizing the process.

$\displaystyle \sum^{\infty}_{k=0}\frac{(-1)^k}{3^k(2k+1)}$
$\displaystyle \sum^{\infty}_{k=0}\left(\frac{1}{3}\right)^k\frac {(-1)^k}{2k+1}$.
I happen to know that,
$\displaystyle \sum^{\infty}_{k=0}\frac{(-1)^k}{2k+1}=\frac{\pi}{4}$ because it is arctangent series.
and also know that,
$\displaystyle \sum^{\infty}_{k=0}\left(\frac{1}{3}\right)^k= \frac{3}{2}$ because it is geometric series. But I do not see how to deal with both of them together.
• Feb 27th 2006, 08:14 AM
CaptainBlack
Quote:

Originally Posted by jallison88
How would one prove that the sum, from n=0 to infinite, of the alternating power series (-1)^n/[3^n*(2n+1)] equals Pi/[2(3)^.5]? I'm having some difficulty conceptualizing the process.

Look at the power series for $\displaystyle \arctan(x)$, then let $\displaystyle x=1/ \sqrt{3}$, and then Bob's your uncle.

RonL