# Thread: Sums and Fourier Series 2

1. ## Sums and Fourier Series 2

Use the answer to the Fourier series:
for

Compute sum:

[(-1)^n] / (2n+1)

Some starting points would be greatly appreciated!!

2. Try putting $\displaystyle x=\pi/2$.

3. Originally Posted by Opalg
Try putting $\displaystyle x=\pi/2$.
Thank you, I've tried this. However, I still don't see where to get
SUM 1/(2n+1)

I think I am missing a fundamental step. Any ideas?

4. Originally Posted by Chitownmegs
Thank you, I've tried this. However, I still don't see where to get
SUM 1/(2n+1)

I think I am missing a fundamental step. Any ideas?
$\displaystyle \sin\pi/2 = 1,\ \sin2\pi/2=0,\ \sin3\pi/2=-1,\ \sin4\pi/2=0,\ \sin5\pi/2=-1$, and in general $\displaystyle \sin(2m\pi/2)=0,\ \sin((2m+1)\pi/2) = (-1)^m.$

Therefore in the sum $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}n\sin(n\pi/2)$, only the odd-numbered terms are nonzero. So try putting n=2m+1, with the sum going from m=1 to ∞.