9th Putnam 1949

a_n is a sequence of positive reals. Show that lim sup_n→∞( (a₁ + a_n+1)/a_n)ⁿ ≥ e.

Solution

It is sufficient to show that (a₁ + a_n+1)/a_n ≥ 1 + 1/n for infinitely many n. For these n then form a sequence for which ( (a₁ + a_n+1)/a_n )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 + a_n+1)/a_n < 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 + a_N+1/N+1 < a_N/N, a₁/N+2 + a_N+2/N+2 < a_N+1/N+1, a₁/N+3 + a_N+3/N+3 < a_N+2/N+2, ... . Hence a_n/N > a₁(1/N+1 + 1/N+2 + ... + 1/N+M) + a_N+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
jscholes@kalva.demon.co.uk
11 Mar 2002