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If $a_k \to 0$ then $\frac{1}{n} \sum_{k=1}^{n} a_k \to 0$ can be established using cauchy theorem on limits, which states that,

if $\lim_{n \to \infty} l_n = l$ then $\lim_{n \to \infty}\frac{1}{n} \sum_{k=1}^{n}a_k = l$.

Converse of Cauchy theorem on limits does not hold in general [a counterexample would be $\{(-1)^n\}$].

So how do you prove that

If $\frac{1}{n} \sum_{k=1}^{n} a_k \to 0$ then $a_k \to 0$ provided that $(a_k) \in (0, 1)$

spaceman_spiff
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  • more http://math.stackexchange.com/questions/1571297/cesaro-means-convergence?rq=1 – Nosrati Jan 18 '17 at 21:48
  • No, this is different: it's going in the other direction. – Robert Israel Jan 18 '17 at 21:51
  • Hi Spaceman - notice that your counterexample (while otherwise valid) does not obey the restriction "$(a_k)$ [is] a sequence in (0,1)". – Chris Jan 18 '17 at 21:53
  • $a_n = \mathbf{1}[n\text{ is a square}]$ has zero Cesaro mean. – Sangchul Lee Jan 18 '17 at 21:57
  • Yeah, @SangchulLee, but the problem there is that that sequence isn't contained in the open interval $(0,1)$. (The issue mainly being the many $0$-terms.) – Chris Jan 18 '17 at 22:00
  • @Chris Ah, we need a minor modification, then. Thankfully Robert Israel showed a way. – Sangchul Lee Jan 18 '17 at 22:02
  • I'll edit the question..... My counterexample is for Cauchy theorem on limits and not for the proposed problem – spaceman_spiff Jan 18 '17 at 22:08
  • Followup question: If $(a_k)$ is a convergent sequence of real numbers such that $\lim_{n \to \infty}\frac{1}{n} \sum_{k=1}^{n}a_k = 0$, must $\lim_{n \to \infty}a_k = 0$? – quasi Jan 18 '17 at 22:10
  • Should I ask the above as my own question, or is it OK as a comment? – quasi Jan 18 '17 at 22:13
  • Never mind -- it's obvious that the answer to my followup question is "yes". – quasi Jan 18 '17 at 22:17
  • @quasi: I'd say it is obvious that the answer to your last followup question is "no". $\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}a_k = 0$ does not imply that $\lim_{n\to +\infty}a_n$ exists. – Jack D'Aurizio Jan 18 '17 at 22:39
  • @Jack D'Aurizio -- In my followup question, I did specify that the sequence $(a_k)$ converges. – quasi Jan 18 '17 at 22:42
  • All right, but in such a case it is trivial anyway. If $a_n\to c$, then $\frac{A_n}{n}\to c$ too, hence if we know that $\frac{A_n}{n}\to 0$ and $a_n$ is convergent, $a_n\to 0$. – Jack D'Aurizio Jan 18 '17 at 22:45
  • @Jack D'Aurizio -- I did say it was obvious. – quasi Jan 18 '17 at 22:47

2 Answers2

1

For example, try $a_{2^n} = 1/2$, $a_n = 1/2^n$ otherwise.

Robert Israel
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For brevity, let $A_n = \sum_{k=1}^{n}a_n$.

1. $\{a_n\}_{n\geq 1}\text{ infinitesimal}\Rightarrow\{A_n/n\}_{n\geq 1}\text{ infinitesimal}$.
This is kind of trivial: if a sequence is infinitesimal, it is infinitesimal on average, too.
For any $\varepsilon>0$, there is some $N_\varepsilon$ ensuring $|a_n|\leq\varepsilon$ as soon as $n\geq N_\varepsilon$.
It follows that if $n=kN_\varepsilon$ with $k>1$ we have $$|A_n| \leq \left|\sum_{k=1}^{N_\varepsilon}a_n\right|+(k-1)N_\varepsilon \varepsilon $$ so $\frac{A_n}{n}$ is bounded by $2\varepsilon$ in absolute value as soon as $n$ is big enough. $\varepsilon$ is arbitrary and we are done.
We do not need the non-negativity of $\{a_n\}_{n\geq 1}$, just its bounded-ness, following from the $\text{infinitesimal}$ assumption.

2. $\{A_n/n\}_{n\geq 1}\text{ infinitesimal}\Rightarrow\{a_n\}_{n\geq 1}\text{ infinitesimal}$.
This is simply false, even with the assumption $a_n\in (0,1)$. It is enough to consider the sequence given by $a_n=\frac{1}{n}$ if $n$ is not a power of $2$ and $\frac{1}{2}$ if $n$ is a power of $2$. This sequence is infinitesimal on average but not infinitesimal tout court. This converse of 1. requires stronger assumptions on the behaviour of $\{a_n\}_{n\geq 1}$ or $\{A_n/n\}_{n\geq 1}$. Hardy's inequality gives a glance at the possible additional assumptions we may require to make 2. hold. As remarked by quasi in the comments, "$\{a_n\}_{n\geq 1}$ is convergent" is one of them, but "$\{A_n/n\}_{n\geq 1}\in\ell^p$ for some $p>1$" works just as well.

Jack D'Aurizio
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