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Let $(a_i)_{i \in \mathbb{N}}$ be a sequence of positive reals such that $$ \limsup_{i \rightarrow \infty} a_i \, i =0. $$ Is this condition necessary and sufficient for $\sum\limits_{i=1}^\infty a_i < \infty$?

Of course, if $a_i = 1/i$, then the series is infinite and if $a_i = (1/i)^{1 + \varepsilon}$, for some $\epsilon >0$, then it is finite, but it is not clear to me what happens if I choose a sequence which decays faster than $1/i$ but not faster than $(1/i)^{1 + \varepsilon}$ for arbitrary small $\varepsilon>0$.

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    $\begingroup$ For the latter, consider the two series $$\sum_{n=2}^\infty\frac{1}{n\log n}\qquad\text{and}\qquad\sum_{n=2}^\infty\frac{1}{n\log^2n}.$$ The left-hand sum diverges still while the right-hand sum converges. $\endgroup$ – Clayton Aug 8 '18 at 19:25
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    $\begingroup$ Incidentally $\limsup a_i/i=0$ is the same as $\lim a_i/i=0$ $\endgroup$ – zhw. Aug 8 '18 at 19:53
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Note that $\sum_{n=2}^\infty\frac1{n\log n}$ diverges (this follows from the integral test), but $\lim_{n\to\infty}n\frac1{n\log n}=0$.

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  • $\begingroup$ Sorry, I actually meant $a_i \, i$ instead of $a_i / i$. $\endgroup$ – QuantumLogarithm Aug 13 '18 at 19:19
  • $\begingroup$ @QuantumLogarithm Then you should edit your question, right?! $\endgroup$ – José Carlos Santos Aug 13 '18 at 19:21
  • $\begingroup$ @QuantumLogarithm I have answered to the new version of the question. $\endgroup$ – José Carlos Santos Aug 13 '18 at 19:28
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It's obviously necessary: if $\sum_na_n<\infty$, then $a_n\to0$, so $a_n/n\to0$.

It is really far from sufficient: take $a_n=\sqrt{n}$; then $a_n/n\to0$, while the series diverges spectacularly.

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