Series problem! Can someone give me a counterexample? Suppose positive series $\sum a_n<+\infty$, does this implies that
    $$\lim_{n\to\infty}na_n=0 .$$
 A: Let $a_n=\frac1n$ whenever $n$ is a positive perfect square, and $a_n=0$ otherwise. Then $\sum_{n\geq1}a_n=\sum_{m\geq1}a_{m^2}=\sum_{m\geq1}\frac1{m^2}=\frac{\pi^2}6<\infty$ but $\lim_{n\to\infty}na_n$ does not exist. If you don't like the terms $a_n=0$ in a positive series, replace them with something like $\exp(-n)$ that decreases sufficiently rapidly so as to not perturb the convergence of the series.
A: No, you can take $a_{n}=\begin{cases}2^{-k}&\textrm{if }n=2^k\textrm{ for some }k\in\mathbb N_0,\\2^{-n}&\textrm{otherwise.}\end{cases}$
Here $2^ka_{2^k}=1$, so the limit doesn't exist, but the series converges anyway, because the geometric series does.
A: Recall the convergence test: $\sum a_n$ and $\sum b_n$ are given positive series, and $\lim_{n\to\infty}\frac{a_n}{b_n}=L$.


*

*If $L=0$ then when $\sum b_n$ converges so does $\sum a_n$; 

*if $L=\infty$ then when $\sum b_n$ diverges so does $\sum a_n$;

*if $L\in(0,\infty)$ then $\sum b_n$ converges if and only if $\sum a_n$ converges.


Note that $na_n = \dfrac{a_n}{\frac1n}$. 
So this limit $\lim_{n\to\infty} na_n$ is the comparison between $\sum a_n$ and $\sum\frac1n$. Since we know that $\sum a_n$ converges and $\sum\frac1n$ diverges to infinity, the limit cannot be nonzero (if it exists, as Dejan Govc's answer show).
