I have just started learning the basics of Fourier series and have some doubts about it. I am aware that Fourier series can be used to compute infinite sums. For example, $\zeta(2)$ and $\eta(2)$ can be evaluated by using the Fourier series expansion of $x^2$, where $x\in[-\pi, \pi]$. $$x^2=\frac{\pi^2}{3}+\sum_{n \ge 1}\frac{4(-1)^n}{n^2}\cos{nx}$$ Letting $x=\pi$ and $x=0$ will yield the required results. This then brings me to my question. Given a sum to compute, how does one determine the appropriate $f(x)$ and $L$? For example, given a sum like $$\beta(3)=\sum_{n \ge 0}\frac{(-1)^n}{(2n+1)^3}$$ , may I ask how we are supposed to know which function we have to consider?

Also, I am interested in knowing how to apply this technique to evaluate sums of the more general form $$\sum_{n \ge 0}\frac{z^n}{(n+a)^s}$$ i.e. the lerch transcendent, and how to determine if it is not possible to utilise this method. (For example, it does not work on $\zeta(2n+1)$)

Thank you for putting up with my ignorance. Help will be greatly appreciated.

  • $\begingroup$ Given that there's no known closed form for your $\beta(3)$, I suspect that there's no known Fourier series that could be used to sum it either. The difficulty is that what you're interested is the inverse problem for Fourier series (e.g. what function generates these Fourier components); that's tractable numerically, but analytically I expect it's much less so. $\endgroup$ Aug 13, 2014 at 15:47
  • $\begingroup$ @Semiclassical Thank you for input. If I am not wrong however, $\beta(3)=\frac{\pi^3}{32}$, and I believe it can be done using Fourier series expansions. Did you mean $\beta(2)$ has no closed form instead? $\endgroup$ Aug 13, 2014 at 22:51
  • $\begingroup$ Ack, you're quite right. Objection withdrawn! $\endgroup$ Aug 13, 2014 at 22:57
  • $\begingroup$ @Semiclassical Thank you for the bounty. I really appreciate it. $\endgroup$ Oct 18, 2014 at 0:55
  • $\begingroup$ @Semiclassical: I derive a recursion for $\beta(2n+1)$ in this answer. $\beta(2n)$ is a harder nut to crack, like $\zeta(2n+1)$. $\endgroup$
    – robjohn
    Oct 21, 2014 at 1:55

1 Answer 1


Here are three ways to compute $\beta(3)$.

In this answer, starting with $\beta(1)=\frac\pi4$, I derive the recursion for $n\gt0$: $$ \beta(2n+1) = -\sum_{k=1}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)\tag{1} $$ from which we get $$ \beta(3)=\frac{\pi^3}{32}\tag{2} $$

We can also compute $\beta(3)$ using contour integration: $$ \begin{align} 0=\frac1{2\pi i}\oint\frac{\pi\csc(\pi z)}{\left(z+\frac12\right)^3}\,\mathrm{d}z &=2\sum_{k=0}^\infty\frac{(-1)^k}{\left(k+\frac12\right)^3}+\operatorname*{Res}_{z=-1/2}\frac{\pi\csc(\pi z)}{\left(z+\frac12\right)^3}\\ &=16\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}-\operatorname*{Res}_{z=0}\frac{\pi\sec(\pi z)}{z^3}\\ &=16\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}-\operatorname*{Res}_{z=0}\frac{\pi\left(1+\pi^2z^2/2+O(z^4)\right)}{z^3}\\ &=16\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}-\frac{\pi^3}2\tag{3} \end{align} $$ Therefore, $$ \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32}\tag{4} $$

If we integrate your equation for $x^2$, we get $$ \frac13x^3=\frac{\pi^2}{3}x+\sum_{n=1}^\infty\frac{4(-1)^n}{n^3}\sin(nx)\tag{5} $$ and plugging in $x=\frac\pi2$, we get $$ \frac{\pi^3}{24}=\frac{\pi^3}6-\sum_{k=0}^\infty\frac{4}{(2k+1)^3}(-1)^k\tag{6} $$ from which we get $$ \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32}\tag{7} $$

  • $\begingroup$ +1 Thank you for sharing a diversity of methods to evaluate $\beta(3)$. It seems that the third method can be generalised to evaluate $\beta(2n+1)$ and $\zeta(2n)$. Do you mind if I ask if a similar methodology (to the third method) can be employed to compute similar sums such as $\displaystyle\sum^\infty_{n=0}\frac{1}{(3n+1)^3}$ ? (which can of course be evaluated via other means such as using the polygamma reflection and addition formulae). Thanks a lot. $\endgroup$ Oct 21, 2014 at 11:30
  • $\begingroup$ Nicely done. Is there some way to link the last two methods explicitly? I have in mind the following observations: 1) The Fourier coefficients can be interpreted as coefficients of some Laurent series for a function analytic on the unit circle. 2) The Dirichlet series can be interpreted as a residue sum of a function like $f(z)\cot z$. So both cases yield a complex-analysis meaning of the sum. What I wonder is if they're really distinct or if one can be mapped onto the other... $\endgroup$ Oct 21, 2014 at 13:51
  • $\begingroup$ The first two methods are related because they both rely on the Laurent expansion of $\sec(\pi z)$. The last method seems to use that $$\mathrm{Im}\left(\mathrm{Li}_3\left(-z\right)\right) =\frac1{12}\arg(z)^3-\frac{\pi^2}{12}\arg(z)$$ when $|z|=1$. $\endgroup$
    – robjohn
    Oct 21, 2014 at 17:52
  • $\begingroup$ Intriguing. So whatever link there is between the method of the first two and that of the third is not entirely obvious. Interesting that the third method, which is so common in a Fourier analysis course, at its core relies to an altogether obscure-looking relationship. $\endgroup$ Oct 22, 2014 at 0:43
  • 1
    $\begingroup$ @Semiclassical: I will look into it some more to see if there is something I missed, but the connection does not appear obvious now. $\endgroup$
    – robjohn
    Oct 22, 2014 at 0:46

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