Question I am trying to do: Prove that if $(x_n)$ is a convergent sequence, then the sequence given by the averages $$ y_n=\frac{x_1+x_2+\dots+x_n}{n} $$ also converges to the same limit.
In the solutions for this exercise, they prove this in a different way, but my approach is this and I want to see if it is correct (my confidence in analysis is low). If not correct, please tell me what I have done wrong:
Assume $(x_n)\to{x}\in\Bbb{R}$
Let $\epsilon_i>0$ be given $\forall{i}\in\{1,2,\dots,n\}$. Let $N_i\in\Bbb{N}$ be s.t. for all $n>N_i$:$$ |x_1-x|<\epsilon_1, |x_2-x|<\epsilon_2|,\dots,|x_n-x|<\epsilon_n $$
Define $\epsilon:=\max\{\epsilon_1,\dots,\epsilon_n\}$
We now write $$y_n=\frac{(x_1-x)+(x_2-x)+\dots+(x_n-x)}{n}+x$$
Choose $N:=\max\{N_1,N_2,\dots,N_n\}$. Given $n>N\ge{N}_i \quad \forall{i}$,
$$|y_n-x|=\left|\frac{(x_1-x)+\dots+(x_n-x)}{n}\right|\le\frac{|x_1-x|+\dots+|x_n-x|}{n}<\frac{\sum_{i=1}^n\epsilon_i}{n}\le\frac{n\epsilon}{n}=\epsilon$$
where the final inequality comes from the definition of $\epsilon$ as the max of all $\epsilon_i$