I was asked this question in a PhD interview. They didn't say whether $a_n$ is non-increasing or strictly decreasing. I tried various examples such as $a_n=\frac 1{n^p}$ where $p \gt 1$, $a_n=\frac 1{n^2 \ln n}$. Then $n a_n \to 0$.

I couldn't answer this question then.

But when I saw Robert Israel's answer to this question, motivated from it, I defined a sequence $a_n$ as below :

$ a_n = \begin{cases} \frac 1n, & \text{if $n$ is a power of 2} \\ 0, & \text{otherwise.} \end{cases}$

Here, $\sum a_n=\frac 12+ \frac 14+ \frac 18+ \cdots =\frac {\frac 12}{1 -\frac 12}=1$ and $\{n a_n\}$ diverges, because there are infinitely many terms of $0$ and $1$ in it. i.e.

$n a_n = \begin{cases} 1, & \text{if $n$ is a power of 2} \\ 0, & \text{otherwise.} \end{cases}$

So $\{na_n\}$ converges in some cases and diverges in some cases. Am I right?

Note : I have gone through these posts, 1,2, 3 too but they have some extra conditions or specific details.


  • $\begingroup$ Makes sense to me. $\endgroup$ May 20 '17 at 8:15
  • $\begingroup$ Yes you are right (and posting a duplicate, for reasons hard to fathom). $\endgroup$
    – Did
    May 20 '17 at 8:48
  • $\begingroup$ Thanks for assuring me about my approach @Did and Jair Taylor. To Did, I respect your decision of making this post duplicate to the post whose answer motivated me. But I just want to state the differences between that post and mine : 1)My post has proof verification tag. I just wanted a feedback on my approach. Also that post doesn't have this tag. 2)(Minor difference) My post has $a_n \ge 0$ as a condition in the hypothesis but that post has condition in hypothesis as $a_n \gt 0$. $\endgroup$
    – Error 404
    May 20 '17 at 9:15
  • $\begingroup$ IOW, the answer of the question you are duplicating is proving more than you... More generally, it is rarely appropriate to post a new question to ask about very minor points of already posted answers. Comments are made for that, among other things. $\endgroup$
    – Did
    May 20 '17 at 9:24
  • $\begingroup$ @Did Okay I see. Next time I will keep this in mind. $\endgroup$
    – Error 404
    May 20 '17 at 9:31