“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
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
(for the Lebesgue measure), and the Volterra operator T:
which is a linear operator acting on H, in fact a Hilbert-Schmidt operator with kernel
which is bounded, and therefore certainly belongs to the space
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
for an arbitrary choice of orthonormal basis (fn)_n of H; (iii) as the integral
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:
For the special case n=0, we have
For non-zero n, we have
and therefore (Parseval, if you wish, or direct computation):
Summing over all n and identifying the two expression for the trace, we get
and hence — unsurprisingly, I presume — we get
(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
for k> 0 integer, using Tk instead of T; it also hints quite strongly that the result will be of the type
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…