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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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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??

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.

$$

Let $\epsilon >0$

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

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

Exists $n_2 \in \mathbb{N}$ 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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