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.

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- Feb 26th 2006, 11:32 AMjallison88[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 AMThePerfectHackerQuote:

Originally Posted by**jallison88**

Quote:

Originally Posted by**jallison88**

$\displaystyle \sum^{\infty}_{k=0}\left(\frac{1}{3}\right)^k\frac {(-1)^k}{2k+1}$.

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 AMCaptainBlackQuote:

Originally Posted by**jallison88**

RonL