İf $\lim x_{n}=x$ then how to show this equation?

Question

If $\lim_{n \to \infty} x_{n}=x$ equality is true how to show this equations? $\lim_{n \to \infty}\dfrac {x_{1}+x_{2}+\ldots +x_{n}} {n}=x$

Answer 1

Let $\epsilon >0$

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

$|\frac{x_1+...+x_n}{n}-x|=|\frac{(x_1-x)+...(x_{n_1-1}-x)}{n}+\frac{(x_{n_1}-x)+...+(x_n-x)}{n}| \leqslant \frac{|x_1-x|+...+|x_{n_1-1}-x|}{n}+ \frac{|x_{n_1-1}-x|+...+|x_n-x|}{n}$

Exists $n_2 \in \mathbb{N}$ such that $$\frac{|x_1-x|+...+|x_{n_1-1}-x|}{n}< \epsilon, \forall n \geqslant n_2$$

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

$|\frac{x_1+...+x_n}{n}-x| \leqslant \epsilon+ \frac{(n-n_1) }{n}\epsilon<2 \epsilon $