Bill Cordwell has called my attention to a paper by Robin Chapman that collects 14 proofs that zeta(2) [:= sum(1/n^2)] = (pi^2)/6. http://www.maths.ex.ac.uk/~rjc/etc/zeta2.ps (or .pdf) Proof #13 is what I asked for in HAKMEM: The proof squares the double-ended series ... + 1/-3 - 1/-1 + 1/1 - 1/3 + 1/5 - ... (easily seen to be just 2*(pi/4) when summed from the center) and shows equality to the double-ended series ... + 1/9 + 1/1 + 1/1 + 1/9 + 1/25 + ... (which is absolutely convergent) and is equal to (pi^2)/4. From there, it's a quick wave of the hands to get zeta(2) = (pi^2)/6. Chapman credits the proof to the Borweins' book "Pi and the AGM", and, indeed, it is exercise 11 in section 11.3, on page 381. The proof only requires some algebra, the logN estimate for a harmonic sum, and limits. The first move is to clip the series, summing each series from the -(2K+1) term through the 2K+1 term. The term squares cancel, leaving the problem of showing that the cross terms in the squared series sum to a value that approaches 0 as K->infinity. Since the clipped series are finite, there are no nasty worries about conditional convergence. It's a little complicated, but a nice first principles proof. Even the logN estimate is dispensable, being replaceable with an upper bound of, say, sqrtN for the harmonic sum. Rich