Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series).

I have , somehow, to find the sum of $\sum_{n=1}^\infty \frac{1}{n^4}$ using Parseval's theorem.

I tried some things that didn't work so I won't post them.

Can you please explain me how do I find the sum of this series using Parseval's identity?


  • 1
    $\begingroup$ I suppose you know this, but you should find a function whose Fourier coefficients are ~ $\frac{1}{n^2}$ $\endgroup$ – Poppy Jan 25 '14 at 15:14
  • 1
    $\begingroup$ You have it solved here: math.cmu.edu/~bobpego/21132/fourierexample.pdf $\endgroup$ – Poppy Jan 25 '14 at 15:21

Let $f(x)=x^2$ for $x\in(-\pi,\pi)$. Computing the Fourier coefficients gives

$$a_n=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 e^{i n x} dx=\frac{2 \cos(\pi n)}{n^2}=2\frac{(-1)^n}{n^2}$$

for $n\in\mathbb{Z}$, $n\not=0$, and $a_0=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 dx=\frac{\pi^2}{3}$.

Therefore $|a_n|^2=\frac{4}{n^4}$ for $n\in\mathbb{Z}$, $n\not=0$ and $|a_0|^2=\frac{\pi^4}{9}$.

By Plancherel/Parseval's theorem,

$$\frac{\pi^4}{9}+8\sum_{n=1}^\infty \frac{1}{n^4}=\sum_{n=-\infty}^\infty |a_n|^2=\frac{1}{2\pi}\int_{-\pi}^\pi x^4 dx=\frac{\pi^4}{5}$$

Simplifying, this gives

$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{8}\left(\frac{1}{5}-\frac{1}{9}\right)=\frac{\pi^4}{90}$$

  • $\begingroup$ Thank you very much .. one question: why is $n$ starts at $-\infty$ and not at $1$? $\endgroup$ – Billie Jan 25 '14 at 15:36
  • 1
    $\begingroup$ You need to consider all Fourier coefficients, that includes those with negative indices. $\endgroup$ – J.R. Jan 25 '14 at 15:37
  • 1
    $\begingroup$ You are welcome. $\endgroup$ – J.R. Jan 25 '14 at 15:40
  • $\begingroup$ @YourAdHere what is the motivation behind choosing $f(x)= x^2$ ? Poppy wrote that we need to find a function whose Fourier coefficients are ~ $\frac{1}{n^2}$. But how are we suppose to guess these functions? $\endgroup$ – Dark_Knight Sep 3 '16 at 1:27
  • 1
    $\begingroup$ @Dark_Knight Monomials, that is $x\to x^k$ are among the most simple functions that one can write down. Now it only remains to notice that the Fourier coefficients of $x^k$ are roughly $1/n^k$ (in magnitude) since we integrate by parts $k$ times to evaluate $\int^{\pi}_{-\pi} x^k e^{inx} dx$. $\endgroup$ – J.R. Sep 9 '16 at 14:39

We have $f \in L^2 \left[ -\pi, \pi \right]$ then \begin{align} \dfrac{1}{2}{A_0^2} + \displaystyle \sum_{n=1}^{\infty}{\left( A_n^2 + B_n^2 \right)} = \dfrac{1}{\pi} \int_{-\pi}^ \pi {f^2(x)} dx. \end{align} Here $A_n$ and $B_n$ are Fourier coefficients of $f$.

Let $f(x) = x^2$ for $x \in \left( -\pi, \pi \right)$. Then $f \in L^2 \left[ -\pi, \pi \right]$ and $$f(x) \sim \dfrac{{\pi}^2}{3} + 4{\displaystyle \sum_{n =1}^{\infty}{ \dfrac{(-1)^{n}}{n^2}\cos nx}}.$$ Therefore $$\dfrac{1}{2}\left(\dfrac{2 \pi^2}{3}\right)^2 +\displaystyle\sum_{n=1}^{\infty} \left( \dfrac{4.(-1)^n}{n^2} \right)^2 = \dfrac{1}{\pi} \int_{-\pi}^\pi{x^4}dx $$

So $$ \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^4}= \dfrac{\pi^4}{90}.$$


Hint: Look at $f(x) = x^2$ on an interval around $0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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