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For all $i\in\mathbb{N}$, let $(a_{i,n})_{n\in\mathbb{N}}$ be a real sequence that tends to $0$ for $n\rightarrow\infty$. It holds also that $|a_{i,n}|\leq1$ for all $i,n\in\mathbb{N}$. Is it possible to show that \begin{align*} c_n:=\frac{1}{n}\sum_{i=1}^{n}a_{i,n}\xrightarrow{n\rightarrow\infty}0?\end{align*} Thanks!

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Counter-example: $$a_{i,n}=\left\{\begin{array}{ll} 1&n< 2i\\0 & n\ge 2i\end{array}\right.\ .$$ It is easy to see that $\lim\limits_{n\to\infty}c_n=\frac{1}{2}$.

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  • $\begingroup$ Can you expand on your "easy to see" statement? Unless I am much mistaken, this sequence has no limit, and is a canonical example of such a sequence... $\endgroup$ – user1729 Jul 17 '13 at 12:40
  • $\begingroup$ @user1729: Can you evaluate $c_n$ using the definitions? $\endgroup$ – 23rd Jul 17 '13 at 15:11
  • $\begingroup$ Oh, sorry, had miss-read what you wrote. $\endgroup$ – user1729 Jul 17 '13 at 15:35
  • $\begingroup$ @user1729: Please never mind. :) $\endgroup$ – 23rd Jul 17 '13 at 15:39
  • $\begingroup$ Thanks a lot for your quick answer! $\endgroup$ – stroem Jul 18 '13 at 7:46
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I think the answer depends on the sequences themselves. Here is what I have thought.

$\forall\ i\in \mathbb{N},\ \forall \epsilon>0,\ \exists N_i\in \mathbb{Z}^+$ s.t. $$|a_{i,n}|<\epsilon$$ for all $n\geq N_i$. Then let $N=\max_{i\in \mathbb{N}}N_i$. Now $\mathbf{IF}\ N<\infty$ then we see that $$|\frac{1}{n}\sum_{i=1}^na_{i,n}|\leq\frac{1}{n}\sum_{i=1}^n |a_{i,n}|<\epsilon$$ for all $n\geq N$, and consequently the sequence $c_n$ converges to $0$. But it does not seem obvious that the $\mathbf{IF}$ always holds.

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  • $ a_{i,n} ≤ 1 $
  • $ ∑a_{i,n} ≤ n $
  • $ (∑a_{i,n}) /n ≤ 1 $
  • $ lim(∑a_{i,n}/n) \rightarrow0$

finaly $C_n \rightarrow0$

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  • $\begingroup$ I do not understand when you say "$C_{n+1} = C_n + \delta(n)\rightarrow0$ but not to $0$." Two $0$s? $\endgroup$ – user1729 Jul 17 '13 at 12:23
  • $\begingroup$ :) thanks user1729 I don't know how to write math on internet. I mean delta(n) ->0 not Cn+delta(n) please help me for editing lol $\endgroup$ – Adrien Jul 17 '13 at 12:23
  • $\begingroup$ Click on "edit" to see what I did. It is LaTeX. There is a tutorial page here. Basically, wrap everything in dollar signs, sub-scripts are "_" while super-scripts (powers) are "^". Use the curly brackets "{" and "}" to group symbols together when applying sub- and super-scripts. So, for example, $e^{i\pi}$ is given by e^{i\pi} rather than e^i\pi or e^(i\pi). These give $e^i\pi$ or $e^(i\pi)$ respectively. (But use normal brackets in your working, for example e^{(i\pi)} gives $e^{(i\pi)}$.) $\endgroup$ – user1729 Jul 17 '13 at 12:27
  • $\begingroup$ (Also, be careful with your dollar signs - you've let one slip from in front of the \delta.) $\endgroup$ – user1729 Jul 17 '13 at 12:31
  • $\begingroup$ @Adrien how do you say $C_2=a_1+a_2/2$? First of all $C_1=a_{1,1}$, then $C_2=\frac{1}{2}(a_{1,2}+a_{2,2})$ and so on. $\endgroup$ – Samrat Mukhopadhyay Jul 17 '13 at 12:53

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