A convergence problem in real non-negative sequence, $\sum_{n=1}^\infty a_n$.

Question

We now have $a_n\geq 0$, $\forall n=1,2,...,$ and $\sum_{n=1}^\infty a_n <\infty$. Then I guess that $\lim_{n\to\infty} a_n \cdot n = 0$. But I realized that it is wrong. Since if we let $a_n = 1/n $ if $n = 2^i$ for some $i=1,2,...$ and $a_n = 0$ for the rest of the $n$.

Then we have that $\sum_{n=1}^\infty a_n = 1/2 + 1/4 + 1/8 + \cdots < \infty$, and $a_n\cdot n$ does not converges to $0$.

Now I add another condition that $a_n$ is non-increasing. Does this result hold this time. i.e. the formal question is as follows:

$a_n\geq 0$, $\forall n=1,2,...,$ and $a_n$ is non-increasing, and $\sum_{n=1}^\infty a_n <\infty$. Then prove $a_n \cdot n \to 0$, or give a counterexample that $a_n\cdot n$ does not necessarily converge to $0$

How is your example a counterexample to the claim? It looks like you're taking $a_n= \frac{1}{2^n}$, which does satisfy $\lim a_n n=0$ — M10687, Aug 16 '16 at 03:46
The counterexample is $a_n = 1/n$ when $n = 2^k$ for some $k \in \mathbb{N}$, and $a_n = 0$ otherwise. The zeroes have been omitted when taking the sum. — Shagnik, Aug 16 '16 at 04:23

Shagnik · Accepted Answer · 2016-10-13T17:37:30.107

4

It is true that $a_n \cdot n \rightarrow 0$ under this condition.

To prove this, observe that if the series is convergent, the sequence of partial sums is a Cauchy sequence. In particular, for every $\varepsilon > 0$, there is some $N_0 = N_0(\varepsilon)$ such that whenever $n > m > N_0$, $\left| \sum_{i=1}^n a_i - \sum_{i=1}^m a_i \right| < \varepsilon$.

Now consider taking $m = N_0 + 1$ and $n \ge 2m$. Then we have $$ \sum_{i=1}^n a_i - \sum_{i=1}^m a_i = \sum_{i=m+1}^n a_i = a_n \cdot (n-m) + \sum_{i=m+1}^n (a_i - a_n) \ge a_n \cdot (n-m),$$ by the non-increasing property of the sequence $(a_n)_n$.

Hence $a_n \cdot (n-m) < \varepsilon$ whenever $n \ge 2(N_0 + 1)$. However, since $n \ge 2m$, $n-m \ge \frac12 n$, and so $a_n \cdot n \le 2 a_n \cdot (n-m) < 2 \varepsilon$.

Since $\varepsilon$ can be taken to be arbitrarily small, this, coupled with the non-negativity of $a_n$, proves $a_n \cdot n \rightarrow 0$.

edited Oct 13 '16 at 17:37

answered Aug 16 '16 at 03:24

Shagnik

3,613

You're welcome, and nice question! – Shagnik Aug 16 '16 at 04:26
@Schrödinger'sCat Do you have any reasons not to accept this?... – drhab Oct 20 '16 at 08:24
@drhab Sorry that I did not realise that I should accept the answer. I did not notice it before. Now I did it. Thanks for your remind. – Schrödinger's Cat Oct 20 '16 at 20:44
@Schrödinger'sCat Ego te absolvo. Nice habit formation took place! – drhab Oct 21 '16 at 07:28

Zau · Answer 2 · 2016-08-18T07:30:18.390

1

Define $ b_n:= n$

$\frac{1}{b_n} \sum_{i = 1}^n b_i a_i = \frac{1}{n} \sum_{i = 1}^n i a_i \geq^{(1)} \frac{1}{n} \sum_{i = 1}^n i a_n \\\ = a_n \frac{1}{n} \sum_{i = 1}^n i = a_n \frac{1}{n} \frac{(n+1)n}2 = a_n \frac{1}{n} \frac{(n+1)n}2 = a_n \frac{n+1}{2} \geq 0$

By Kronecker's lemma, $\lim_{n\to \infty}\frac{1}{b_n} \sum_{i = 1}^n b_i a_i = 0$

By squeezee theorem, $ 0 = \lim_{n\to \infty} \frac{1}{b_n} \sum_{i = 1}^n b_i a_i \geq \lim_{n\to \infty} a_n \frac{n+1}{2} \geq 0$

Hence, $\lim_{n\to \infty} a_n (n+1) = 0$

By the convegence of summation of $a_n$, $\lim_{n \to \infty} -a_n = 0$.

Therefore, $\lim_{n\to \infty} a_n n = \lim_{n\to \infty} a_n (n+1) - a_n = 0 $

Where the (1) comes from the decreasing of the sequence.

edited Aug 18 '16 at 07:30

answered Aug 16 '16 at 03:25

Zau

3,909
12
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But we don't necessarily have $\displaystyle \sum_{i=1}^\infty a_i=0$, I don't see how this implies the result. – M10687 Aug 16 '16 at 03:30
@M10687 Sorry now I fixed it. – Zau Aug 18 '16 at 07:33

A convergence problem in real non-negative sequence, $\sum_{n=1}^\infty a_n$.

2 Answers2