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Can it be proven that : If $a_n>0$ and $\sum a_n$ converges then $na_n \to 0$ (without assuming $a_n$ is decreasing)

Please note This question does not require $a_n$ to be decreasing as condition, Is it possible to prove $na_n \to 0$ without requiring $a_n$ to be decreasing? I keep thinking that $a_n>0$ and $\sum a_n$ converging then that implies that $a_n$ must be decreasing anyway(not monotonically decreasing necessarily )

Is it possible that $a_n>0$ and $\sum a_n$ converges then $na_n \not \to 0$?

There are already answered questions that assume $a_n$ to be decreasing as a requirement: (this question does not assume $a_n$ is decreasing)

If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$

Prove that if $\sum a_n$ converges, then $na_n \to 0$.

This was the post that made me ask this question, a counter example of monotonoic decreasingness of $a_n$ was given in comments by the OP.

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For example, try $$a_n = \cases{ 1/n & if $n$ is a power of $2$\cr 2^{-n} & otherwise} $$

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  • $\begingroup$ How I can show that the series $\sum a_n$ is convergent ? $\endgroup$ – Empty Apr 2 '17 at 6:52
  • $\begingroup$ $\sum a_n=\sum 1/2^k+\sum_{n\neq 2^k} 1/2^n$. $\endgroup$ – checkmath Sep 15 '17 at 23:39

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