Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have found the Fourier Series of $$f(x)= \begin{cases} 0\colon & -\pi<x<0 \\ x\colon & 0<x<\pi \end{cases}$$ to be equal to: $$f(x)=\frac{\pi}{4}+\sum_{n=1}^\infty \left(\frac{1}{\pi n^2}((-1^n)-1)\cos(nx)+\frac{(-1)^{n+1}}{n}\sin(nx)\right)$$ I am not quite sure though how to use that to evaluate: $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$ I know the answer is supposed to be $\frac{\pi^2}{8}$. Should I change the index in the series to match that in the expression and nullify $\sin(nx)$? Would Parseval's identity be of use here?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

You have incorrectly calculated value of "an" instead of pi^2 , it comes out to be only "pi". And then put x=0 and take the LHL and RHL limit of the f(x) to get the value of f(0), which comes out to be "0". Then we see that sin nx =0 i.e. term "bn"=0 and only "an" term remains. Please click on the link to see the solution. https://drive.google.com/file/d/0B5FWdKs3BipkdVBteTBWTEZmNkk/edit?usp=sharing


share|improve this answer
I cannot see anything when I click your link. –  Grtv Jan 21 '14 at 19:31
Please see now drive.google.com/file/d/0B5FWdKs3BipkdVBteTBWTEZmNkk/… if u dont see anything the download the picture by going to 'File' menu –  anjan Jan 21 '14 at 19:35

Try plugging in $x = 0$. On one hand, it is a theorem that the Fourier series of a piecewise smooth function (such as the one you describe) converges to the average of the left limit and the right limit of the function at every point. On the other hand, you can explicitly plug in $x = 0$ into the series. This gives:

$$0 = \frac{\pi}{4} + \sum_{n = 1}^\infty \frac{1}{n^2 \pi^2}\left((-1)^n - 1\right)$$

since $\cos(0) = 1$ and $\sin(0) = 0$. If you re-index the sum on the right and rearrange things, you should get what you want.

share|improve this answer
Below is an attempt: $$0 = \frac{\pi}{4} + \sum_{n = 0}^\infty \frac{1}{(2n+1)^2 \pi^2}(-2)$$ Hence, that sum should be equal to $\frac{\pi^3}{8}$. What am I doing wrong? –  Grtv Jan 21 '14 at 19:22
I inadvertently typed $\pi^2$ instead of $\pi$. Thanks a lot! :-) –  Grtv Jan 21 '14 at 19:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.