# Is the series with positive terms convergent if $na_n \rightarrow 0$?

I am unsure how to tackle this problem:

If $\{a_n\}$ is a sequence of positive numbers with $\lim_{n\rightarrow \infty } na_n=0$, what can you conclude about the convergence of $\sum a_n$? (*)

First of all, the book I'm referring to proved that if the terms of the convergent series $\Sigma a_n$ decrease monotonically then $\lim_{n\rightarrow \infty } na_n=0$. As such I was led to the intuition that my problem, (*), implies divergence. I could not, however, find a counterexample to prove my point and it's something I suck at, immensely.

Nevertheless, I attempted on "some proof" of which I am unsure of:

If $\lim_{n\rightarrow \infty } na_n=0$ then $\forall \epsilon>0 \,\,\exists N$ s.t. $na_n < \epsilon \,\, \forall n\geq N$. As the terms are positive, $0<a_n \leq na_n < \epsilon$. Thus, $\lim_{n\rightarrow \infty} a_n = 0$.

Here I'm stuck and I'm hoping someone can guide me to the right path...

I still have a gut feel that the series is divergent, I just can't find a counterexample. I would also appreciate if anyone can give me some hints as to how to construct counterexamples

Help very much appreciated

• No, if $p_n$ is the $n$th prime, then $\sum \dfrac{1}{p_n}$ is divergent, but $\lim \dfrac{n}{p_n} = 0$. Possibly easier example is to show that $\sum \dfrac{1}{n\log n}$ is divergent. – Thomas Andrews Jul 18 '13 at 17:51
• (Of course, $\dfrac{1}{n\log n}$ has to start at $n=2$. But you can figure that part out.) – Thomas Andrews Jul 18 '13 at 17:58
• "what can you conclude about the convergence"... Short answer: nothing, the series can converge and it can diverge. – Did Jul 18 '13 at 18:27

The basic example of a divergent series with $na_n\to 0$ is $a_n=\dfrac{1}{n\log n}$, for $n\geq 2$. Proving this diverges is related to the fact that $$\int_{2}^n \frac{1}{x\log x} dx = \log \log n - C$$ and $\log\log n\to\infty$ as $n\to\infty$.