The real sequence an satisfies an = ∑n+1∞ ak2. Show ∑ an does not converge unless all an are zero.
Clearly an ≥ 0. If any an = 0, then all subsequent ai must be zero, and, by a trivial induction, all previous ai. So assume no an = 0.
Notice that an+1 = an2 + an. But we have just shown that an2 > 0, so an+1 > an.
If the sum converges, then we can take n sufficiently large that an+1 + an+2 + an+3 + ... < 1. Then an = an+12 + an+22 + an+32 + ... < an(an+1 + an+2 + an+3 + ... ) < an. Contradiction. So the sum does not converge.
14th Putnam 1954
© John Scholes
26 Nov 1999