Prove that if $\lim_{n \to \infty}z_{n}=A$ then: $$\lim_{n \to \infty}\frac{z_{1}+z_{2}+\cdots + z_{n}}{n}=A$$ I was thinking spliting it in: $$(z_{1}+z_{2}+\cdots+z_{N-1})+(z_{N}+z_{N+1}+\cdots+z_{n})$$ where $N$ is value of $n$ for which $|A-z_{n}|<\epsilon$ then taking the limit of this sum devided by $n$ , and noting that the second sum is as close as you wish to $nA$ while the first is as close as you wish to $0$. Not sure if this helps....

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    $\begingroup$ Possibly easier to first show for $A=0$ $\endgroup$ – Thomas Andrews Oct 5 '12 at 19:19
  • $\begingroup$ @Thomas Andrews I edited the question, adding my idea. Could you tell me what you think, please? $\endgroup$ – Mykolas Oct 5 '12 at 19:29
  • $\begingroup$ Changed the title. There is no series here. $\endgroup$ – Did Oct 5 '12 at 20:54
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    $\begingroup$ Maybe this question was asked here before (I did not search), but at least I remember that we had questions such that this can be obtained as a consequence of the results from those questions, see, e.g., If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\limsup\sigma_n \leq \limsup s_n$ and limit of quotient of two series. $\endgroup$ – Martin Sleziak Oct 6 '12 at 9:28
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    $\begingroup$ It is also worth mentioning that this is often called Cesaro mean. $\endgroup$ – Martin Sleziak Oct 27 '12 at 6:53

It seems like Homework problem, hence I'll just give hint: $$\frac{z_1+z_2+\cdots +z_n}{n}-A=\frac {(z_1-A)+(z_2-A)+\cdots +(z_n-A)}{n}$$ Now use the defn of limit that for every $\epsilon > 0$ there exists $N_0 \in \mathbb N$ such that $|z_m-A| < \epsilon \ \forall m \geq N_0$

Also remember triangle inequality : $|a_1+a_2+\cdots +a_n| \leq |a_1| + |a_2| +\cdots +|a_n|$

Can you find proper $a_i$ in terms of say $z_i$'s??

  • $\begingroup$ thankyou for help. It's not homework. Would $a_{i}$ be $\frac{z_{i}}{n}$, and since for $n \to\infty $all sums go to $0$, it would be proved? @TheJoker $\endgroup$ – Mykolas Oct 5 '12 at 19:40
  • $\begingroup$ and would my first idea be wrong? $\endgroup$ – Mykolas Oct 5 '12 at 19:44
  • $\begingroup$ You can take $a_i=\frac{z_1-A}{n}$. Now use the fact that limit of $z_i$ is $A$. about your approach,it is true,just write rigorously :-).also there is no dofference between mine and your approach as your next step would be same. $\endgroup$ – TheJoker Oct 5 '12 at 19:51

Let $\epsilon >0$

We have that $\lim_{n \rightarrow \infty}z_n=A$ thus $\exists n_1 \in \mathbb{N}$ such that $|z_n-A|< \epsilon, \forall n \geqslant n_1$

$|\frac{z_1+...+z_n}{n}-A|=|\frac{(z_1-A)+...(z_{n_1-1}-A)}{n}+\frac{(z_{n_1}-A)+...+(z_n-A)}{n}| \leqslant \frac{|z_1-A|+...+|z_{n_1-1}-A|}{n}+ \frac{|z_{n_1}-A|+...+|z_n-A|}{n}$

Exists $n_2 \in \mathbb{N}$, by Archimedean property, such that $$\frac{|z_1-A|+...+|z_{n_1-1}-A|}{n}< \epsilon, \forall n \geqslant n_2$$

Now for $n \geqslant n_0= \max\{n_1,n_2\}$

$|\frac{z_1+...+z_n}{n}-A| \leqslant \epsilon+ \frac{(n-n_1) }{n}\epsilon<2 \epsilon $

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    $\begingroup$ Thanks for your answer; could you please explain why 'there exists $n_2$ such that...' $\endgroup$ – Daniele1234 May 20 '18 at 21:58
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    $\begingroup$ Because the numerator is a positive constant number and we use the Archimedean property of the real numbers(or you can see it as the convergence of $1/n$ to zero) $\endgroup$ – Marios Gretsas May 21 '18 at 1:01

This can be an easy consequence of a more general statement which is from Polya's Problems and Theorems in Analysis:

Let $\{a_n\}_{n=1}^{\infty}$ be a real sequence such that $\lim_{n\to\infty}a_n=a$. And we have a family of finite sequences $\{\{b_{nm}\}_{m=1}^{m=n}\}_{n=1}^{\infty}$: $$ b_{11}\\ b_{21},b_{22}\\ b_{31},b_{32},b_{33}\\ \cdots $$ such that $$ b_{mn}\geq 0 $$ for all $m,n$, and $\sum_{m}b_{nm}=1$ for each $n=1,2,\cdots$. Let $\{c_n\}_{n=1}^{\infty}$ be such that $$ c_n=\sum_{m=1}^na_mb_{nm} $$ Then $\lim_{n\to\infty}c_n=a$ if and only if $\lim_{n\to\infty}b_{nm}=0$ for each $m$.

The question in OP is a special case of the statement by letting $$ b_{nm}=\frac{1}{n},\quad m=1,2,\cdots. $$


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