Question

By using both sides of the identity \[\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\], find the sum of the series \[\sum_{n=0}^{\infty}\frac{4(-1)^n}{2n+1}\]

Answer 1

I get to the point where I have \[\frac{1}{2}(\ln(1+x)-\ln(1-x))+C=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\] which seems close enough to where I need to go? but I don't know what to do next

Answer 2

look at the arctangent

Answer 3

\[\tan^{-1}(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}\]

Answer 4

\[\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}=\cdots\]

Answer 5

oh wait is it if we substitute x=1, we get rid of the annoying 2n+1 power, so arctan(1)= \[\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\]
and so to get the 4, we do 4*arctan(1),
and arctan(1) = pi/4,
so 4*arctan(1) = pi??

Answer 6

yes

Answer 7

ahh I see :)! I didn't think of using arctan ._. so thanks!

Answer 8

no problem