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Let $ \Sigma_{n=1}^\infty a_n$ be a non-negative series, I need your help please with proving/disproving that:

if $\lim_{n \to\infty} n \cdot a_n = 1/2 $ then $\Sigma_{n=1}^\infty a_n$ is divergant.

I tried to disprove with a counterexample by taking $a_n = 1/n^2$ but then the first condition isn't true.

It seems like in order for the first condition to be true, $a_n$ must be of the form $a_n = 1/n$ and then it seems like the condition is true but I'm not sure how to prove it.

Thanks

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  • $\begingroup$ You have $a_n\sim 1/(2n)$ as $n\to +\infty$, so... $\endgroup$ – Rigel May 20 '17 at 14:25
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Hint: there is $n_0\in\mathbb{N}$ such that for each $n\ge n_0$, $na_n>\frac{1}{4}$. So $a_n>\frac{1}{4n}$ for $n\ge n_0$.

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  • $\begingroup$ now I see it. solved. thank you. $\endgroup$ – Noam May 20 '17 at 14:29

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