Putnam 1973/A2 solution

34th Putnam 1973

Problem A2

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

Solution

Answer: yes.

Consider a_n + a_n+1 + ... + a_n+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 s_n be the sum of the first n terms. We have shown that with an unequal number of positive and negative signs s_8n diverges and hence s_n does not converge. With an equal number of signs, s_8n converges. But each term tends to zero, so |s_8n+i - s_8n| tends to zero for i = 1, ... 7. Hence s_n converges.

34th Putnam 1973