34th Putnam 1973

Problem A2

an = ±1/n and an+8 > 0 iff an > 0. Show that if four of a1, a2, ... , a8 are positive, then ∑ an converges. Is the converse true?



Answer: yes.

Consider an + an+1 + ... + an+7. It is ( ±(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7) ±n(n+2)(n+3)(n+4)(n+5)(n+6)(n+7) ± ... ± n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6) ) / ( n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7) ). If there are an equal number of plus and minus signs, then there is no n7 term in the numerator, so the expression is O(1/n2). On the other hand if there are an unequal number of plus and minus signs, then there is an n7 term in the numerator and the expression is O(1/n). But ∑ 1/n diverges, and ∑ 1/n2 converges.

Let sn be the sum of the first n terms. We have shown that with an unequal number of positive and negative signs s8n diverges and hence sn does not converge. With an equal number of signs, s8n converges. But each term tends to zero, so |s8n+i - s8n| tends to zero for i = 1, ... 7. Hence sn converges.



34th Putnam 1973

© John Scholes
22 Aug 2001