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We have a non-negative sequence $\{a_n\},~n\in\mathbb{N}$ and its arithmetic mean $\frac{\sum_{i=0}^{N-1}a_i}{N}$ goes to 0 as $N\to\infty$. Can we claim that in this case, there exists a subsequence $\{a_{k_n}\}$ such that $\lim_{n\to\infty}a_{k_n}=0$?

I have a proof of yes as follows:

We prove it by contradiction. Suppose that such a subsequence does not exist. Then there exists an integer $N$ and positive constant $\epsilon$ such that $\forall n\ge N$, $a_n\ge\epsilon$. However, in this case, $\lim_{N\to\infty}\frac{\sum_{i=0}^{N-1}a_i}{N}=\epsilon>0$, which contradicts the fact that $\lim_{N\to\infty}\frac{\sum_{i=0}^{N-1}a_i}{N}=0$.

My question is: “Suppose that such a subsequence does not exist. Then there exists an integer $N$ and positive constant $\epsilon$ such that $\forall n\ge N$, $a_n\ge\epsilon$.” Is this statement correct?

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Yes, your statement is correct if you mean an infinite subsequence:

Suppose that such an infinite subsequence does not exist. Then there exists an integer $N$ and a positive constant $\epsilon$ such that, for all $n>N$, we have $a_n>\epsilon$.

The hypothesis that $a_n\ge0$ is important. Without it, there are counterexamples to the theorem.

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  • $\begingroup$ As far as I recall, a subsequence in mathematics is always defined to be infinite. $\endgroup$ – Clement C. Jan 13 '18 at 4:08
  • $\begingroup$ @Clement Yes! But Google the words "finite sequence" and "finite subsequence" - you might be surprised! :)) $\endgroup$ – Alex Jan 13 '18 at 5:31

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