Follow-up Question about Cesaro mean proof

1. The first part is controlled because $N_1$ is fixed and the denominator $n$ can be arbitrarily large.
2. For the second part you are using the estimate $|x_n-x|<\epsilon$ for large $n$. Note that there are totally $n-N_1$ terms in the numerator, the absolute value of each is less than $\epsilon$.

To summarize, you are first given $\epsilon>0$. Then you have $N_1$ such that $n\ge N_1$ implies $|x_n-x|<\epsilon$, in particular $$ |x_{N_1+1}-x|<\epsilon,\quad |x_{N_1+2}-x|<\epsilon,\cdots,|x_{n}-x|<\epsilon\tag{1} $$ which implies by the triangle inequality that $$ \left|\frac{(x_{N_1+1}-x)+\cdots+(x_{n}-x)}{n}\right|<\frac{(n-N_1)\epsilon}{n}\tag{1'} $$

Then, for this (fixed) $N_1$, since $$ \lim_{n\to\infty}\frac{|(x_1-x)+\cdots+(x_{N_1}-x)|}{n}=0 $$ there exists $N_2$ such that $n\ge N_2$ implies $$ \frac{|(x_1-x)+\cdots+(x_{N_1}-x)|}{n}<\epsilon\tag{2} $$

So if you pick $N_3=\max(N_1,N_2)$, then $n\ge N_3$ implies both (1') and (2).

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If you can show that for every $\epsilon>0$, there exists $N>0$ such that $n\ge N$ implies $|a_n|<2\epsilon$, it follows that $$ \lim_{n\to\infty}a_n=0 $$ You don't have to have exactly $\epsilon$ in the estimate.

first red text: You want to make a seperate statement about the first $N_1-1$ terms since you don‘t know how the sequence behaves in those terms. You know that $|x_n-x|<\varepsilon$ for all $n\geq N_1$ but you don‘t know what‘s going on earlier in the sequence.

second red text: this follows again from the triangle inequality: $$ \left| \frac{x_{N_1}+x_{N_1+1}+\ldots+x_n}{n} - x \right| = \left| \frac{x_{N_1}-x+x_{N_1+1}-x+\ldots+x_n-x}{n} \right| \leq \frac{|x_{N_1}-x|+|x_{N_1+1}-x|+\ldots+|x_n-x|}{n} $$

that are $N_1-n$ terms of value $<\varepsilon$ (that’s because of how we chose $N_1$) in the numerator giving you the term $$ \frac{n-N_1}{n}\varepsilon $$

edit: regarding the term $2\varepsilon$: it is only important that we can make this expression arbitrarily small. Although it is „convention“ to use $\varepsilon$ it is also fine to use $2\varepsilon$ since we can make this expression as small as we want if we just make $\varepsilon$ small enough.