11th Putnam 1951 Problem A3 Find ∑ 0 ∞ (-1) n /(3n + 1).
We have ln(1 + x) = x - x 2 /2 + x 3 /3 - x 4 /4 + ... (1). Put ω = e 2πi/3 , so that ω 3 = 1 and 1 + ω + ω 2 = 1. Then ω 2 ln(1 + ωx) = x - ωx 2 /2 + ω 2 x 3 /3 - x 4 /4 + ... (2) and w ln(1 + ω 2 x) = x - ω 2 x 2 /2 + ωx 3 /3 - x 4 /4 + ... (3). Adding (1), (2), (3) we get 3(x - x 4 /4 + x 7 /7 - ... ) = ln(1 + x) + ω 2 ln(1 + ωx) + ω ln(1 + ω 2 x). Hence the required series sums to 1/3 (ln 2 + ω 2 ln(1 + ω) + ω ln(1 + ω 2 ) ). If k = ln(1 + ω), then e k = 1 + ω = - ω 2 = e iπ/3 , so k = iπ/3. Hence ω 2 ln(1 + ω) = -(1 + i√3)/2 iπ/3 = π/2√3 - iπ/6. Similarly, ω ln(1 + ω 2 ) = -iπ/3 (-1/2 + i√3/2) = π/2√3 + iπ/6. Hence the series sums to 1/3 ln 2 + π/3√3. 11th Putnam 1951 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002