Fascinating! I had a different objective in mind when I wrote the problem. My goal was to reach the fairly difficult pi^2/6 result using series manipulation, starting from a definition of pi based on 1 - 1/3 + 1/5 - ... or the like. Ideally, one would avoid calculus and complex integrations and the theory of dilogarithms, or Euler's trick with the power series for sin x. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Robert Baillie Sent: Wednesday, February 22, 2006 8:39 AM To: math-fun Subject: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8 HAKMEM 239 has this interesting question: "119 (Schroeppel): Can someone square some series for Pi to give the series Pi^2/6 = 1 + 1/2^2 + 1/3^2 + ... ?" Would you settle for squaring a series for Pi/Sqrt(8) to get a series for Pi^2/8? Pi/Sqrt(8) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 --++ ... Pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9+2 + 1/11^2 + ... Also, notice that you can square the first series by adding the squares of the individual terms! I first noticed this by comparing exercises 6a and 6b on page 398 in Wilfred Kaplan's "Advanced Calculus". The series for Pi/Sqrt(8) comes from the Fourier series for f1(x) = 0 for -Pi <= x < 0, f(x) = 1 for 0 <= x <= Pi. The series is f[x_] := (1/2) + (2/Pi) Sum[ Sin[(2n-1)x]/(2n-1), {n, 1, Infinity}] f[Pi/4] gives the series for Pi/Sqrt[8]. The series for Pi^2/8 comes from the Fourier series for g1(x)= Abs[x], -Pi <= x <= Pi. g[x_] := (Pi/2) - (4/Pi) Sum[ Cos[(2n-1)x]/((2n-1)^2), {n, 1, Infinity}] g[0] gives the series for Pi^2/8. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun