Follow-up Question about Cesaro mean proof I am trying to understand the proof behind the Cesaro mean converging. I am using https://math.stackexchange.com/a/2342856/633922 (hopefully it is also correct) as a guide because it seems very direct. I will comment on the steps I understand and where I need help.
The statement: If $(x_n)$ converges to $x$, the sum of averages $y_n=\dfrac{x_1+x_2+\cdots+x_n}{n}$ also converges to the same limit.
Proof:
Since $(x_n)$ converges, given an arbitrary $\epsilon >0$, there exists an $N_1\in\mathbb{N}$ such that whenever $n\geq N_1$ we have $|x_n-x|<\epsilon$.
(Definition of convergent sequence)
Now,
$$\begin{align*}
\left|\frac{x_1+x_2+\cdots+x_n}{n}-x\right|=&\left|\frac{(x_1-x)+\cdots+(x_{N_1-1}-x)}{n}+\frac{(x_{N_1}-x)+\cdots+(x_{n}-x)}{n}\right|\\
\leq& \left|\frac{(x_1-x)+\cdots+(x_{N_1-1}-x)}{n}\right|+\left|\frac{(x_{N_1}-x)+\cdots+(x_{n}-x)}{n}\right|\text{ (Triangle inequality)}
\end{align*}
$$
Now we want to make a statement about the first $N_1-1$ terms, $\color{red}{why?}$ That is:
By the Archimedean principle we can find an $N_2$ such that whenever $n\geq N_2$ we have that
$$\left|\frac{x_1+x_2+\cdots+x_{N-1}}{n}\right|<\epsilon
$$
(Thought: is it because $x_1,\dots,x_{N_1-1}$ is finite?)
Now we can choose an $N_3=\max\{N_1,N2\}$ such that for all $n\geq N_3$ we have (My thought: Is this because choosing the max of both will always guarantee the final inequality to always work?)
$$
\left|\frac{x_1+x_2+\cdots+x_n}{n}-x\right|\leq \underbrace{\epsilon}_{N_1-1}+\underbrace{\color{red}{\frac{n-N_1}{n}}}_{\text{why and how?}}\epsilon< 2\epsilon
$$
And this finishes the proof. I always assumed the ending statement has to be (something)$<\epsilon$ or is this saying that each sum of the right side of the triangle inequality is less than $\epsilon/2$. I would really appreciate the help on the areas I am thoroughly confused about.
 A: *

*The first part is controlled because $N_1$ is fixed and the denominator $n$ can be arbitrarily large.

*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).

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.
A: 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.
