Find ∑0∞ (-1)n/(3n + 1).
Solution
We have ln(1 + x) = x - x2/2 + x3/3 - x4/4 + ... (1). Put ω = e2πi/3, so that ω3 = 1 and 1 + ω + ω2 = 1. Then ω2 ln(1 + ωx) = x - ωx2/2 + ω2x3/3 - x4/4 + ... (2) and w ln(1 + ω2x) = x - ω2x2/2 + ωx3/3 - x4/4 + ... (3).
Adding (1), (2), (3) we get 3(x - x4/4 + x7/7 - ... ) = ln(1 + x) + ω2 ln(1 + ωx) + ω ln(1 + ω2x). Hence the required series sums to 1/3 (ln 2 + ω2 ln(1 + ω) + ω ln(1 + ω2) ).
If k = ln(1 + ω), then ek = 1 + ω = - ω2 = eiπ/3, so k = iπ/3. Hence ω2 ln(1 + ω) = -(1 + i√3)/2 iπ/3 = π/2√3 - iπ/6. Similarly, ω ln(1 + ω2) = -iπ/3 (-1/2 + i√3/2) = π/2√3 + iπ/6. Hence the series sums to 1/3 ln 2 + π/3√3.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002