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....
Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means
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0You might also be interested in the first problem in this problem set from MIT OCW: http://ocw.mit.edu/courses/mathematics/18-100c-real-analysis-fall-2012/assignments/MIT18_100CF12_ps5.pdf – 2013-08-15
3 Answers
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??
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0You 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. – 2012-10-05
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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1Because 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) – 2018-05-21
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. $