a_{n} = ±1/n and a_{n+8} > 0 iff a_{n} > 0. Show that if four of a_{1}, a_{2}, ... , a_{8} 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 n^{7} term in the numerator, so the expression is O(1/n^{2}). On the other hand if there are an unequal number of plus and minus signs, then there is an n^{7} term in the numerator and the expression is O(1/n). But ∑ 1/n diverges, and ∑ 1/n^{2} 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.

© John Scholes

jscholes@kalva.demon.co.uk

22 Aug 2001