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If a series $\sum u_n$ converges, $u_n \to 0$ but is there any type of condition on $u_n$ that would ensure that $nu_n \to 0$?

I can think of many $u_n's$ satisfying this, all geometric series terms, p-series terms too for $p\geq 2$, so maybe there's a general condition on $u_n$ that along with $\sum u_n < \infty$ guarantees that $nu_n \to 0$.

reason I'm asking this is because, reading an article an author makes the following claim in a rather casual manner so I must be missing something.

Claim :

$\Sigma \operatorname{Pr}\left(|X|>n^{1 / r}\right)<\infty$ ensures $n \operatorname{Pr}\left(|X|>n^{1 / r}\right) \rightarrow 0$

$X$ is an arbitrary random variable and $0<r<2$.

the_firehawk
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    That is true if $(u_n)$ is decreasing, see for example https://math.stackexchange.com/q/4603/42969 – Martin R Mar 28 '24 at 10:45
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    And counter example if not decreasing https://math.stackexchange.com/q/2084371/399263 – zwim Mar 28 '24 at 11:14
  • We can prove ... if $\sum u_n$ converges, then $n u_n$ does not converge to any nonzero value. [As noted, $n u_n$ may or may not converge to $0$.] – GEdgar Mar 28 '24 at 15:22

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