# Proving divergence of a limit

$(a_n)_{n\ge 1}$ be a sequence of positive reals such that $a_1+a_2+\cdots +a_n<n^2$ for all $n\ge 1$. Prove that $$\displaystyle \lim_{n\to \infty}\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)=\infty$$ My attempt : Suppose eventually $a_{n+1}<2n+1$ then we have our condition is satisfied but also $$\displaystyle \sum \frac{1}{a_{n+1}}>\sum \frac{1}{2n+1}$$ but the right hand side diverges. Though I am not sure if this process is correct and probably it is not but any and all help will be welcomed. Thanks in advance.

Let $k \in \Bbb{N}$. Then use Cauchy Schwarz to obtain
$$2^k = \sum_{\ell = 2^k}^{2^{k+1}-1} 1 = \sum_{\ell = 2^k}^{2^{k+1}-1} \left[\frac{1}{\sqrt{a_\ell}} \cdot \sqrt{a_\ell}\right] \leq \sqrt{\sum_{\ell = 2^k}^{2^{k+1}-1}\frac{1}{a_\ell}} \sqrt{\sum_{\ell = 2^k}^{2^{k+1}-1} a_\ell}.$$
• For the intuition behind this: You should check that applying the above inequality just on $\ell = 1,\dots n$ does not work. The idea is then that the $a_i$ should be approximately constant on the set $2^k, \dots ,2^{k+1}$ so that the $a_i^{-1}$ are approximately proportional to the $a_i$ so that Cauchy Schwarz is approximately sharp. – PhoemueX May 30 '14 at 16:45