# First evaluation of $\sum_{n \geq 1} 1/n^2$ by Fourier series

There are many ways to evaluate $\sum_{n \geq 1} 1/n^2$ as $\pi^2/6$, including multiple solutions using Fourier series. A colleague asked me who was the first person to use Fourier series (or Fourier analysis) to make this computation. I don't know and was unable to figure this out. Any concrete evidence that a particular person was the first to do this would be appreciated.

• It must have been before Kummer (1846). – user2255 Jun 26 '16 at 18:41
• @FranzLemmermeyer, can you clarify why? For example, did Kummer include such a solution in something he wrote in which it was clear the idea was not due to him? – KCd Jun 26 '16 at 18:53
• If no one comes up with something better I will add details. Kummer discussed more complicated sums in connection with cubic Gauss sums and did not give references. He might have learned this from Dirichlet . . . – user2255 Jun 26 '16 at 19:15

Kummer's interest in this question is due to the problem of the location of cubic Gauss sums, analogous to Gauss's determination of the sign of quadratic Gauss sums. In De residuis cubicis disquisitiones nonnullae analyticae (Some analytic investigations on cubic residues), Crelle 32 (1846), Kummer studies cubic Gauss sums and connects these sums with L-series of cubic characters. He accomplishes this by using Fourier series and a very clever trick - I'm still trying to decipher his ideas, and I don't think this has been taken up by anyone at all. Kummer does not look at all at the simpler case of quadratic Gauss sums, where I think his methods yields a clever evaluation of $L(1,\chi)$ for quadratic characters, let alone at the (rather trivial) case of the zeta function, but it is rather obvious that Kummer regarded these special cases (Leibniz series or Euler's $\zeta(2)$) as less interesting or at least as not novel - at the time he was trying to impress Jacobi and Dirichlet. Kummer does not give any sources, but at the end he remarks that some of his formulas for cubic residues may be proved directly without the analytic methods that he regards as being invented by himself.
• I found the article at digizeitschriften.de/download/PPN243919689_0032/… and see at the start of section 4 (p. 350) that he writes down $\sum_{n \geq 0} \cos((2n+1)v)/(2n+1)^2 = \pi^2/8 - (\pi/4)v$ for $0 \leq v \leq \pi$. Setting $v = 0$ gives $\sum_{n \geq 0} 1/(2n+1)^2 = \pi^2/8$, from which $\sum_{n \geq 1} 1/n^2 = \pi^2/6$ easily follows, although I don't see how Kummer evaluates that Fourier series to be a quadratic polynomial on $[0,\pi]$, and especially (key point) how he finds the constant term is $\pi^2/8$. – KCd Jul 2 '16 at 23:52
• Isn't the whole point of using Fourier series the fact that computing the Fourier expansion of e.g. linear functions such as $f(x) = x$ is a routine calculation? – user2255 Jul 3 '16 at 8:41
• Whoops, I should not have said "quadratic" polynomial. And now I see what you mean. If we start with the simple even triangular wave $f(x)=1−(2/\pi)|x|$ for $−\pi\leq x\leq\pi$, which has value $1$ in the middle, $−1$ at the endpoints, and repeats periodically outside $[-\pi,\pi]$, then $f(x)$ is continuous everywhere and its Fourier series is easily found to be $\sum_{n\geq 0} (8/(\pi^2(2n+1)^2))\cos((2n+1)x)$. Setting $x=0$, we get $\sum_{n \geq 0} 1/(2n+1)^2 = \pi^2/8$. The series Kummer wrote down is $(\pi^2/8)f(x)$ on $[0,\pi]$, but it's easier to imagine someone had worked out (contd.) – KCd Jul 3 '16 at 14:46
• the Fourier series of the even triangular wave with max/min values $\pm 1$ and then found the sum of reciprocal odd squares and the expression $\pi^2/8$ coming out, although I don't know if that is what actually happened. It's hard to believe anyone would have started from the linear function on $[0,\pi]$ that appears in Kummer's paper and worked out the Fourier series of that. – KCd Jul 3 '16 at 14:47
• For that matter, a more basic continuous even piecewise linear function is $|x|$ on $[-\pi,\pi]$, repeated periodically. Its Fourier series is $\pi/2 - \sum_{n \geq 0} (4/(\pi(2n+1)^2))\cos((2n+1)x)$. Setting $x = 0$ we get $\pi/2 = \sum_{n \geq 0} 4/(\pi(2n+1)^2)$, so $\sum_{n \geq 0} 1/(2n+1)^2 = \pi^2/8$. It would be good to find out who first developed some continuous even piecewise linear functions into Fourier series. – KCd Jul 3 '16 at 15:06