14th Putnam 1954

The real sequence a_n satisfies a_n = ∑_n+1^∞ a_k². Show ∑ a_n does not converge unless all a_n are zero.

Solution

Clearly a_n ≥ 0. If any a_n = 0, then all subsequent a_i must be zero, and, by a trivial induction, all previous a_i. So assume no a_n = 0.

Notice that a_n+1 = a_n² + a_n. But we have just shown that a_n² > 0, so a_n+1 > a_n.

If the sum converges, then we can take n sufficiently large that a_n+1 + a_n+2 + a_n+3 + ... < 1. Then a_n = a_n+1² + a_n+2² + a_n+3² + ... < a_n(a_n+1 + a_n+2 + a_n+3 + ... ) < a_n. Contradiction. So the sum does not converge.

14th Putnam 1954

© John Scholes
jscholes@kalva.demon.co.uk
26 Nov 1999