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Let $a_n$ be a sequence of positive numbers. Suppose $\limsup(na_n) = 1$. Does this mean $\sum a_n$ diverges?

I have only concluded this if the limit superior is in fact the limit of the sequence. Also, I have managed to prove this if the limit inferior is strictly greater than $0$ (but then it is true for every limit superior)

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  • $\begingroup$ notice you could modify the answer given so that $\limsup n a_n = \infty$ $\endgroup$ – clark Jun 14 '16 at 9:27
  • $\begingroup$ If the sequence is monotone, then convergence implies that $\lim_{n\to\infty} na_n=0$. See here: Series converges implies $\lim{n a_n} = 0$. $\endgroup$ – Martin Sleziak Jun 14 '16 at 10:43
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No. Consider $a_n=1/n$ while $n=2^k$ for some $k$, and $a_n=1/n^2$ while $2^k<n<2^{k+1}$.

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  • $\begingroup$ Why does this series converges? $\endgroup$ – Joshhh Jun 14 '16 at 9:33
  • $\begingroup$ You need to calculate it carefully. $\sum_n a_n<\sum_k 1/2^k+\sum_n1/n^2$ $\endgroup$ – lostlife Jun 14 '16 at 9:34
  • $\begingroup$ OK, so after you posted this, here is my attempt: If $k$ is the greatest such that $2^k < m$, then $S_m = \sum\limits_{j=1}^{k+1}\frac{1}{2^j} + \sum\limits_{j\neq 2^l}\frac{1}{n^2} < \sum\limits_{j=1}^m\frac{1}{2^j} + \sum\limits_{j=1}^m\frac{1}{n^2}$, and the rightmost expression converges, hence from the comparison the series converges. Is this ok? $\endgroup$ – Joshhh Jun 14 '16 at 9:57
  • $\begingroup$ Sorry, the fraction in the second sums is $\frac{1}{j^2}$ $\endgroup$ – Joshhh Jun 14 '16 at 10:03

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