# Putnam 1994, Question 1

Just want to record a solution to a relatively simple Putnam problem here. The problem is the following:

Show that a sequence $a_1,a_2,a_2,\dots$ satisfies the condition $0. Prove that $\sum a_i$ diverges.

I tried attacking it with the usual methods of showing that a sequence diverges, but nothing seemed to work. However, every good Olympiad worth its salt asks for some experimentation, which will then yield important observations/patterns. I did the same here.

Without loss of generality, let $a_1=1$. Then $a_2+a_3\geq a_1=1$. Hence, $a_2+a_3\geq 1$. Similary, $a_4+a_5\geq a_2$ and $a_6+a_7\geq a_3$. Therefore, $a_4+a_5+a_6+a_7\geq a_2+a_3\geq 1$. Continuing this pattern, we observe that for all $n\in\Bbb{N}$, $a_{2^n}+\dots+a_{2^{n+1}-1}\geq 1$. Hence $\sum a_i$, which can be thought of as $\sum\limits_{n=1}^{\infty}(a_{2^n}+\dots+a_{2^{n+1}-1})$, diverges.