convergence of sum of squares over squared sum

Question

If we have $\frac{\sum_{i=1}^{n} a^2_i}{(\sum_{i=1}^{n} a_i)^2}$ where $a_i$ is a positive sequence where each element is finite with probability 1, under which set of conditions does the division converge to $0$ as n approaches infinity?. I am sure the next statement is true: \begin{equation} \frac{\sum_{i=1}^{n} a^2_i}{(\sum_{i=1}^{n} a_i)^2}\leq \frac{\sum_{i=1}^{n} a^2_i}{\sum_{i\neq j} a_ia_j} \end{equation} Is there maybe a way to show the denominator in the RHS grows faster than the numerator under some additional assumptions?

Answer 1

Assuming $\space i \space$ is an integer.

\begin{align*} \dfrac{\sum_{i=1}^{n} a^2_i}{\big(\sum_{i=1}^{n} a_i\big)^2} &= \dfrac{ \dfrac{n(n_+1)(2n+1)}{6} }{ \bigg(\dfrac{n(n+1)}{2}\bigg)^2}\\ \\ &=\dfrac{4n(n_+1)(2n+1) }{ 6n^2(n+1)^2}\\\\ &=\dfrac{2(2n+1)}{3n(n+1)} \end{align*}

Since the numerator is linear and the denominator is quadratic, it appears that the series tends to zero.

Answer 2

Well, here's a condition where it isn't true. If $a_i \ge c a_{i-1}$ for all $i$, where $c > 1$, then $$\sum_{i=1}^n a_i \le \frac{c}{c-1} a_n$$ so $$ \frac{\sum_{i=1}^n a_i^2}{\left(\sum_{i=1}^n a_i\right)^2} \ge \left(\frac{c-1}{c}\right)^2$$ does not converge to $0$.