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If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$

Since $(a_n)$ is decreasing and bounded below, by Monotonic Convergence Theorem, $(a_n)$ converges. So, there exists $N$ such that $|a_n-L|< \epsilon$

Since $\sum{a_n}$ is convergent, by the Divergence test, we have $\lim_n{a_n}=0$, which means there exists $N$ such that $|a_n-0|< \epsilon$

Then I get stuck at here.

I try to figure out the statement's meaning by inserting some examples, like $a_n=\dfrac{1}{n^2}$

Can anyone guide me on this question?


marked as duplicate by Martin Sleziak, Alex Provost, user26857, John B, user228113 Apr 1 '16 at 0:40

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Suppose that $\lim\limits_{n\to\infty}na_n\ne0$. That is, there is an $\epsilon\gt0$ so that for any $N$, there is an $n\ge N$ so that $na_n\ge\epsilon$. Since $a_n$ is decreasing, $$ \sum_{k=n/2}^n a_n\ge\frac{n}{2}\frac{\epsilon}{n}=\frac{\epsilon}{2} $$ Since $na_n$ does not converge to $0$, we can find infinitely many such blocks from $n/2$ to $n$ that sum to at least $\epsilon/2$. Thus, we've shown the contrapositive: if $\lim\limits_{n\to\infty}na_n\ne0$, then $\sum\limits_{k=0}^\infty a_n$ diverges.

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    $\begingroup$ How did you show $\lim_{n\to\infty}na_n$ exist? $\endgroup$ – ziyuang Feb 10 '16 at 21:17
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    $\begingroup$ I assumed that the limit was not $0$ and got a contradiction. Part of the limit not being $0$ is the limit not existing. Thus, the contradiction implies not only that the limit exists, but also that the limit is $0$. $\endgroup$ – robjohn Feb 11 '16 at 0:24
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    $\begingroup$ That answer is not satisfactory, we can only guarantee n_ja_{n_j}\geq \epsilon. $\endgroup$ – checkmath Sep 15 '17 at 23:21
  • $\begingroup$ @checkmath: That is all one needs to deny convergence. $\endgroup$ – robjohn Sep 16 '17 at 2:32

Hint Use Cauchy's criterion for the convergence of a series. Remember the sequence is decreasing and positive.

  • $\begingroup$ Do you mean the Cauchy condensation test? $\endgroup$ – Jack Aug 13 '13 at 18:31
  • $\begingroup$ @Jack No. I mean Cauchy's criterion: a sequence of reals converges iff it is Cauchy. $\endgroup$ – Pedro Tamaroff Aug 13 '13 at 18:57
  • $\begingroup$ Ah, I see. So it's in @robjohn's answer, right? $\endgroup$ – Jack Aug 14 '13 at 1:30
  • $\begingroup$ @Jack I am suggesting you prove $p\to q$, not $\neg q\to\neg p$, but the idea is the same. $\endgroup$ – Pedro Tamaroff Aug 14 '13 at 1:32

Given $\epsilon >0$, there exists an $N$ large enough such that $n,m\ge N \implies \sum a_k =|\sum a_k|<\epsilon$. Then for any $n\ge N$ we have that $$0\le na_{2n}\le a_n+a_{n-1}+ \ ... \ +a_{2n-1}=\sum a_k <\epsilon,$$ thus we have $\lim na_{2n}=0$.

Now fix $\epsilon >0$ and fix $N$ such that $n\ge N \implies na_{2n}=|na_{2n}|<\epsilon/3$. Now for $n>2N$ we can find an $m\ge n$ such that $n/3 \le m\le n/2$, hence we have $|ma_{2m}| <\epsilon/3$. Since $a_k$ is a decreasing sequence then we have that $|ma_n|\le |ma_{2m}|<\epsilon/3$, thus $|na_n|\le |3ma_n|<\epsilon$, since $n\le 3m$. Therefore $\lim na_n=0$.

Sorry if this is a bit messy. If you want a different proof or want me to add more detail, let me know.

  • $\begingroup$ It should be $m\geq N$. $\endgroup$ – Kaa1el Feb 11 '14 at 17:04
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    $\begingroup$ I would simplify the end, since you have that $2na_{2n}\to 0$ we only need to prove that $(2n+1)a_{2n+1}\to 0$ but that holds since $0\leq (2n+1)a_{2n+1}\leq (2n+1)a_{2n}=2na_{2n}+a_{2n}$ which go both to zero. $\endgroup$ – checkmath Sep 15 '17 at 23:49

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