an is a sequence of positive reals. Show that lim supn→∞( (a1 + an+1)/an)n ≥ e.
It is sufficient to show that (a1 + an+1)/an ≥ 1 + 1/n for infinitely many n. For these n then form a sequence for which ( (a1 + an+1)/an )n has a limit point (possibly infinity) not less than lim (1 + 1/n)n = e.
If that were false, then we would have (a1 + an+1)/an < 1 + 1/n for all n not less than some N. In particular, taking n = N, N+1, N+2, ... we get a1/N+1 + aN+1/N+1 < aN/N, a1/N+2 + aN+2/N+2 < aN+1/N+1, a1/N+3 + aN+3/N+3 < aN+2/N+2, ... . Hence an/N > a1(1/N+1 + 1/N+2 + ... + 1/N+M) + aN+M/N+M > a1(1/N+1 + 1/N+2 + ... + 1/N+M). But this holds for any M and (1/N+1 + 1/N+2 + ... ) diverges. Contradiction.
9th Putnam 1949
© John Scholes
11 Mar 2002