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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.

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  • $\begingroup$ Perhaps $a_n=\frac{1}{n}$? $\endgroup$ – Hypernova Nov 28 '19 at 16:42
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    $\begingroup$ @Hypernova a_n is a sequence of positive integers. $\endgroup$ – Sayan Nov 28 '19 at 16:43
  • $\begingroup$ 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. $\endgroup$ – Greg Martin Nov 28 '19 at 16:44
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$a_n = \frac1n$, $b_n = 1$, $\frac{b_n}{\sum_{i = 1}^n a_i} \rightarrow 0$, $\frac{b_n^2}{\sum_{i = 1}^n a_i^2} \rightarrow\frac{6}{\pi^2}$.


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.

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  • $\begingroup$ Ah I see! I was too occupied with straight forward polynomial type sequences. It makes sense. Thanks a lot! $\endgroup$ – Sayan Nov 28 '19 at 16:59

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