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, 2014 at 15:14
  • 2
    $\begingroup$ You have it solved here: math.cmu.edu/~bobpego/21132/fourierexample.pdf $\endgroup$
    – Poppy
    Jan 25, 2014 at 15:21

4 Answers 4


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, 2014 at 15:36
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    $\begingroup$ You need to consider all Fourier coefficients, that includes those with negative indices. $\endgroup$
    – J.R.
    Jan 25, 2014 at 15:37
  • 1
    $\begingroup$ You are welcome. $\endgroup$
    – J.R.
    Jan 25, 2014 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? $\endgroup$ Sep 3, 2016 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, 2016 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 $0$


The series $\sum_{n\ge 1}\frac{1}{n^4}$ also appears in the following funny way. Consider the operator ${\rm d}^2/{\rm d}x^2$ on $[0,\pi]$ s.t. Dirichlet boundary conditions. Find Green's function $G(x,s)$. Find the normalized in $L^2[0.\pi]$ eigenfunctions $u_n(x)$ of the operator; the eigenvalues are $\lambda_n=n^2$.

Evaluate the integral $\int_0^\pi\int_0^\pi G^2(x,s){\rm d}x~{\rm d}s$ in two ways, a) using the explicit formula for $G(x,s)$ and b) using the representation $G(x,s)=\sum_{n\ge 1}\frac{u_n(x)u_n(s)}{\lambda_n}$.

After simplifications, you will find $\sum_{n\ge 1}\frac{1}{n^4}=\frac{\pi^4}{90}$.


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