Let an = ∑1n (-1)i+1/i. Assume that limn→∞ an = k. Rearrange the terms by taking two positive terms, then one negative term, then another two positive terms, then another negative term and so on. Let bn be the sum of the first n terms of the rearranged series. Assume that limn→∞ bn = h. Show that b3n = a4n + a2n/2, and hence that h ≠ k.
If we simply write out a4n = (1 - 1/2 + 1/3 - 1/4 + ... - 1/4n) and a2n/2 = (1/2 - 1/4 + 1/6 - 1/8 + ... - 1/4n) and add term by term we find that a4n + a2n/2 = b3n.
Taking the limit, we have immediately that h = 3k/2. But clearly k > 0, since if we group the terms in pairs, each pair is positive. Hence h ≠ k.
14th Putnam 1954
© John Scholes
26 Nov 1999