“Every sum is a trace” is a well-known folklore saying in certain automorphic circles (echoed by a no less convincing “Every sum is an expectation” around probabilists); in this spirit, let’s have a look at one of the most famous sums
$latex \zeta(2)=\sum_{n\geq 1}{\frac{1}{n^2}},$
which Euler was the first to evaluate.
It is possible to see it, and then compute it, as a trace, and I’m sure this has been done many times; here is (a variant of) the way I presented things for an exercise for my Spectral Theory class.
Consider the Hilbert space
$latex H=L^2([0,1])$
(for the Lebesgue measure), and the Volterra operator T:
$latex Tf(x)=\int_0^x{f(y)dy}$
which is a linear operator acting on H, in fact a Hilbert-Schmidt operator with kernel
$latex k(x,y)=\mathbf{1}_{\{y\leq x\}}$
which is bounded, and therefore certainly belongs to the space
$latex L^2([0,1]^2,dxdy).$
The operator T is therefore compact, and the operator S=T*T is also compact, and in fact positive, so that the trace is well-defined as a non-negative real number, or infinity. The trace is well-known to be expressible in different ways: (i) as a sum of the series formed with the eigenvalues of S (with multiplicity); (ii) as the sum of the series
$latex \sum_{n}{\langle S(f_n),f_n\rangle}=\sum_{n}{||Tf_n||^2},$
for an arbitrary choice of orthonormal basis (fn)_n of H; (iii) as the integral
$latex \int_0^1\int_0^1{|k(x,y)|^2dxdy}.$
This last integral is of course completely elementary: it is the area of the lower-half triangle in the square [0,1]2 below the diagonal; in other words, it is 1/2.
For an alternate expression (hence an identity), we look at the series above for the easiest orthonormal basis available:
$latex f_n(t)=e^{2i\pi nt}\quad\quad\text{ for } n\in\mathbf{Z}.$
For the special case n=0, we have
$latex Tf_0(x)=x,$
hence
$latex \langle Sf_0,f_0\rangle =||Tf_0||^2=\int_0^1{x^2dx}=\frac{1}{3}.$
For non-zero n, we have
$latex Tf_n(x)=\frac{e^{2i\pi nx}-1}{2i\pi n},$
and therefore (Parseval, if you wish, or direct computation):
$latex \langle Sf_n,f_n\rangle = ||Tf_n||^2=\frac{1}{4\pi^2 n^2}\times (1+1)=\frac{1}{2\pi^2n^2}.$
Summing over all n and identifying the two expression for the trace, we get
$latex \frac{1}{2}=\frac{1}{3}+\sum_{n\not=0}{\frac{1}{2\pi^2n^2}}=\frac{1}{3}+\frac{1}{\pi^2}\zeta(2),$
and hence — unsurprisingly, I presume — we get
$latex \zeta(2)=\frac{\pi^2}{6}.$
(I said unsurprisingly, but I first managed to get confused enough about the computation — for a slightly different operator — that, for a while, I almost convinced myself that ζ(2)=π2/12).
As a proof (which, I repeat, is certainly not new, though it is not found in this collection), this is fairly close in flavor to the Fourier-expansion proofs, where one expands (typically) the function x-1/2 on [0,1] into Fourier series before applying the Parseval identity. (In fact, it seems this is “dual” in some simple way which could be made precise).
Like the Fourier-expansion argument, it has the nice feature of showing almost immediately that it will be possible to generalize the argument to
$latex \zeta(2k)=\sum_{n\geq 1}{\frac{1}{n^{2k}}}$
for k> 0 integer, using Tk instead of T; it also hints quite strongly that the result will be of the type
$latex \zeta(2k)=\alpha_k\pi^{2k},$
for some rational number αk. But it is equally obvious that this will not work at all like this for zeta evaluated at odd positive integers, as it should…
