$\frac{a_{max}}{\sum a_i} \to 0$: Proof or Counterexample

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.

• Perhaps $a_n=\frac{1}{n}$? Nov 28 '19 at 16:42
• @Hypernova a_n is a sequence of positive integers. Nov 28 '19 at 16:43
• Interesting. If $a_n$ grows like a polynomial in $n$ then both limits are $0$; if $a_n$ grows exponentially in $n$ then both limits are positive. Nov 28 '19 at 16:44

$$a_n = \frac1n$$, $$b_n = 1$$, $$\frac{b_n}{\sum_{i = 1}^n a_i} \rightarrow 0$$, $$\frac{b_n^2}{\sum_{i = 1}^n a_i^2} \rightarrow\frac{6}{\pi^2}$$.
Above was my answer when I didn't notice the requirement that $$a_n$$ are positive integers.
if $$n = 3^k$$ for some integer $$k$$, then $$a_n = 2^k$$; otherwise $$a_n = 1$$.