# If $\limsup(na_n) = 1$, then $\sum\limits_{n=1}^{\infty} a_n$ diverges

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)

• notice you could modify the answer given so that $\limsup n a_n = \infty$ – clark Jun 14 '16 at 9:27
• If the sequence is monotone, then convergence implies that $\lim_{n\to\infty} na_n=0$. See here: Series converges implies $\lim{n a_n} = 0$. – Martin Sleziak Jun 14 '16 at 10:43

No. Consider $a_n=1/n$ while $n=2^k$ for some $k$, and $a_n=1/n^2$ while $2^k<n<2^{k+1}$.
• You need to calculate it carefully. $\sum_n a_n<\sum_k 1/2^k+\sum_n1/n^2$ – lostlife Jun 14 '16 at 9:34
• 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? – Joshhh Jun 14 '16 at 9:57
• Sorry, the fraction in the second sums is $\frac{1}{j^2}$ – Joshhh Jun 14 '16 at 10:03