TITLE: $\frac{a_{max}}{\sum a_i} \to 0$: Proof or Counterexample
QUESTION [7 upvotes]: Suppose I have a sequence of positive integers $\{a_n\}$. Let us denote $b_n=\max_{1\le i\le n} a_i$. Suppose
$$\frac{b_n}{\sum\limits_{i=1}^n a_i} \to 0$$
then show that
$$\frac{b_n^2}{\sum\limits_{i=1}^n a_i^2} \to 0$$
I am not sure if it is true. But I didn't find any Counterexample. I was trying to get a reasonable lower bound for the denominator. I could not find any. Bounds like
$$\sum_{i=1}^n a_i^2 \ge \sum_{i=1}^n a_i$$
won't help though. Note that the converse is true. As you can easily get an upper bound using:
$$\sum_{i=1} a_i^2 \le b_n\sum_{i=1} a_i$$
Any help/suggestions?
Edit: Note that $a_n$'s are positive integers, that's why $\sum a_i^2 \ge \sum a_i$ is true.

REPLY [3 votes]: $a_n = \frac1n$, $b_n = 1$, $\frac{b_n}{\sum_{i = 1}^n a_i} \rightarrow 0$, $\frac{b_n^2}{\sum_{i = 1}^n a_i^2} \rightarrow\frac{6}{\pi^2}$.

Above was my answer when I didn't notice the requirement that $a_n$ are positive integers.
Then a counter example which works with this restriction:
if $n = 3^k$ for some integer $k$, then $a_n = 2^k$; otherwise $a_n = 1$.
The verification is kind of straightforword.