# Series problem! Can someone give me a counterexample?

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

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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

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.

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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).

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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.

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