Demystifying tricks – If $\{x_n\}$ converges, then Cesaro Mean converges.
Exercise 2.3.11 (Cesaro Means). (a) Show if ${x_n}$ is a convergent sequence, then the sequences given by the averages ${\dfrac{x_1 + x_2 + ... + x_n}{n}}$ converges to the same limit.
I rewrote and colored the official solution.
Let $\epsilon>0$ be arbitrary. Then we need to find an $N \in \mathbb{N} \qquad \ni n \geq N \implies\ \frac{x_1 + x_2 + ... + x_n}{n}  L< \epsilon \tag{1}$.
Question posits $(x_{n}) \to L$. So $\exists ; M \in \mathbb{N} \ni n \ge M \implies x_{n}L< M \quad (2)$.
$\text{Also } \exists ; C \ni n \ge C \implies x_{n}L< \epsilon/2. \quad \tag{3}$
My question 1. Where does (3) come from? How to prognosticate $\epsilon/2$? Normally you start with $\epsilon$.
 Doesn't the same argument prove all the terms can be bounded? Why simply 'the early terms in the averages can be bounded' ? PWhy "Because the original sequence is convergent, we suspect that we can bound ... we will be breaking the limit in two at the end" ?
I still don't understand how "we suspect" these bounds?
Now for all $n \ge C$, we can write

How to presage rewriting $\color{red}{L = nL/n}?$ and subtracting $\color{red}{L}?$ from the $x_i$?

How to presage splitting the sum between $x_{C  1}$ and $x_C$?
Now apply the Triangle Inequality to each of the $n$ $(x_i  L)$ terms. Can you please correct the MathJax below? It works on S.E. The second line of inequality is supposed to align with the first. Please remove this once corrected.
$\begin{align} \le & \frac{1}{n}( & \color{green}{\left x_{1}L\right +\ldots +\left x_{c1}L\right} & + \color{brown}{\left x_{C}L\right +\ldots +\left x_{n}L\right} & )\ & & \color{green}{\text{Each of these $(C  1)$ terms < M by (2)} } & \color{brown}{ \text{ Each of these $(n  C)$ terms < $\epsilon/2$ by (3)}} & \ \le & \frac{1}{n}( & \color{green}{(C  1)M} & + \color{brown}{\frac{e}{2}(n  C)} & ). \tag{4}\ \end{align}$
 Why's the last inequality (4) above $\le$? Why not $<$ like (2) and (3)?
Because C and M are fixed constants at this point, we may choose $N_2$ so that $\color{green}{(C  1)M}\frac{1}{n} < e/2 \tag{5}$ for all $n \ge N_2$. Finally, let $N$ [in $(1)$] $= \max{C, N_2}$ be the desired $N$.
 Why are we authorized to choose $N_2$ so that $\color{green}{(C  1)M}\frac{1}{n} < e/2$?
Can you please correct the MathJax below? It works on S.E. The second line of inequality is supposed to align with the first. Please remove this once corrected.
∵ $C \in \mathbb{N} \quad \therefore n  C < n \iff \color{magenta}{\frac{n  C}{n}} < 1 \tag{6}$
$\begin{align} \text{Equation (4) is} & & \color{green}{(C  1)M}\frac{1}{n} & + \frac{e}{2}\color{magenta}{\frac{n  C}{n}}. \ \text{Then by (5) and (6),} & & < e/2 & + e/2\color{magenta}{(1)}/ QED. \ \end{align}$
Stephen Abbott. Understanding Analysis (2016 2 edn). p. 55.
1 answer
There's a lot going on here, so I'll start with some overview. It looks like the idea of the proof is to split $1,...,n$ into two pieces, where the first piece is of fixed size $C$, and the second piece has average close enough to $L$. The first piece has fixed size but the second piece can be made so big that it comprises "almost all" of the average.
You usually end with $\epsilon$, but in more complicated proofs, like this one, you sometimes have to bound two (or more) different terms to get the final sum less than ε. In this case, we have two pieces, one of size $C$ and one of size $nC$, so that usually means we want each piece to be under $\epsilon/2$ so their sum is less than $\epsilon$. This answers (1) and (4).
Now we go after (2). Since $x_n \to L$, $x_n  L \to 0$. So the convergent sequence $x_n  L$ must be bounded by some number $M$. (If you're confused why a convergent sequence must be bounded, you can ask that question separately.) You'll see why we need this later. If you were trying to come up with the proof yourself, you might skip this part for now, and then later realize you don't know anything about the sizes of $x_1$ through $x_{C1}$. Or perhaps you intuitively realize that in an average, any fixed finite number of terms can be made "insignificant"; to make this precise you would use the bound $M$.
Subtracting the $L$ comes about because we already know two inequalities relating to $x_j  L$. It is always less than $M$, and if $j > C$ then it is less than $\epsilon/2$. So if you take the expression for the average, $\left(x_1 + x_2 + ... + x_n)/n  L\right$, there are two pieces with that sum: one from $x_1$ to $x_{C1}$, where each term is within $M$ of $L$, and another from $x_C$ to $x_n$, where each term is within $\epsilon/2$ of $L$. The $nL/n$ is just algebraic manipulation to put a $L$ on each of the $x_j$ terms.
Finally, you ask about this $N_2$. Notice that $(C1)M$ is a fixed constant. So if we divide by a "large enough" $n$, then $(C1)M/n$ can be as small as we want.
(The comment about $\leq$ vs $\lt$ is mostly pedantic; I think you are right that $<$ would be appropriate in the second part.)
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