Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose positive series $\sum a_n<+\infty$, does this implies that $$\lim_{n\to\infty}na_n=0 .$$

share|cite|improve this question
If you add a monotonicity condition, then the limit is true. By the Cauchy Condensation test, since $\sum a_n$ converges, so does $\sum 2^n a_{2^n} $ which implies $2^n a_{2^n} \to 0.$ Now if $ 2^n < k < 2^{n+1} $ then $ka_k < 2^{n+1} a_{2^n} \to 0. $ – Ragib Zaman Apr 10 '12 at 13:03
yes, it is ture, and I know this result! Thank you! – Riemann Apr 10 '12 at 14:45
up vote 15 down vote accepted

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.

share|cite|improve this answer
Thank you very much!I think you are right! – Riemann Apr 10 '12 at 10:00
@Riemann: You're welcome. – Dejan Govc Apr 10 '12 at 10:03

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).

share|cite|improve this answer
but does the limit $\lim na_n$ exist? – Riemann Apr 10 '12 at 9:54
@Riemann: Dejan Govc gave a good counterexample for when the limit does not exist. – Asaf Karagila Apr 10 '12 at 9:57
Will the downvoter explain? – Asaf Karagila Apr 10 '12 at 10:25

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.