Question about 'A convergence sequence of real numbers is bounded.'

Question

The textbook proves this theorem as following (I am writing shortly rather than the exact words):

Let $\lim (x_n)=x$ and $\epsilon:=1$, then there exists $K=K(1)$... (rest of the definition). Then,

$$|x_n|=|x_n-x+x|\le|x_n-x|+|x|\lt1+|x|$$

Let $M:= \sup \{|x_1|,|x_2|, ... , |x_{K-1}|,1+|x|\}$

And the proof follows.

I understand most of the proof, just have trouble understanding, why up-to $x_{K-1}$ ?

My main question is, is the following proof valid? If not, why?

Given sequence is convergent. So, using the usual notations from the definition, the sequence $X=(x_n)$ has a limit $x$, therefore by definition and triangle inequality $|x_n|-|x|\le|x_n-x|<\epsilon$. Which implies, $|x_n|<M, \forall n\in\Bbb{N}$ by letting $M:=\epsilon+|x|$.

Answer 1

It is not valid, since you only have

$$|x_n|-|x|\le|x_n-x|<\epsilon$$

for all $n\ge N$, for some natural number $N = N(\epsilon)$. So

$$|x_n| \le |x|+\epsilon$$

might not be satisfied by $x_1, \cdots, x_{N-1}$.

Answer 2

Your argument is valid only for $n$ sufficiently large. To complete the proof, you have only to observe that the first $n$ elements of the sequence are necessarily bounded, because there are only finitely many of them.