$\max_{k=1}^{n}|x_k| \xrightarrow[\text{}]{\text{$n \rightarrow \infty$}} \sup_{k=1}^{\infty}|x_k|$

Question

I want to prove the following: Let $(x_n)$ be a bounded sequence, then $\max_{k=1}^{n}|x_k| \xrightarrow[\text{}]{\text{$n \rightarrow \infty$}} \sup_{k=1}^{\infty}|x_k|$ .

My Calculations:

Let $c:=\sup_{k=1}^{\infty}|x_k|$. If $c$ is the supremum of $(x_n)$, then

(1) $\forall n \in \mathbb{N}: |x_n| \leq c$

(2) $\forall \epsilon>0 :\exists |x_j|$ such that $|x_j|>c-\epsilon$

Just to clarify it, I am looking at the sequence $(|x_n|)_{n \in \mathbb{N}}$ and NOT at the sequence $(x_n)_{n \in \mathbb{N}}$. This is the reason why I get absolute values in (1) and (2).

Now considering the inequality in (2):

$|x_j|>c-\epsilon$ , we get

$\epsilon > c- |x_j|$

I noticed $\epsilon$ and $c -|x_j|$ are both positive numbers. This means I can "take the absolute value" of the inequality and get $|c- |x_j||<\epsilon $

The inequality I am looking for is $|c-\max_{k=1}^n|x_k|| <\epsilon$.

Now there can be two cases:

Case 1:

$|x_j|$ is the maximum, i.e $|x_j|=\max_{k=1}^n|x_k|$ and $|x_{n+1}|>|x_j|$ But this just means the inequality still holds. Since $|x_{n+1}|>|x_j|$ it follows that $|c-|x_j||<|c-|x_{n+1}||<\epsilon$

Case 2:

$|x_j|$ is the maximum, i.e $|x_j|=\max_{k=1}^n|x_k|$ and $|x_{n+1}|<|x_j|$ In this case nothing changes, since $|x_j|=\max_{k=1}^{n+1}|x_k|$ and the equality $|c- \max_{k=1}^{n+1}|x_k||=|c-|x_j||<\epsilon $ still holds.

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By this reasoning it follows that, for every $n \in \mathbb{N}$, there exists a $\epsilon>0$ such that: $|\underbrace{sup_{k=1}^{\infty}|x_k|}_{=c} - \max_{k=1}^n|x_k||<\epsilon$.

The question I would like to ask is: Is this proof right?

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Edit:

One user pointed out that I did not show the definiton of convergence. In fact, I messed up a little bit in the last part. So now the second try:

So I looked up the definition. Let $y_n$ be a sequence and $y \in \mathbb{R}$. The sequence is convergent if for every $\epsilon>0$ , there exists a $N \in \mathbb{N}$ such that $|y_n-y|<\epsilon$ for all $n>N$

So let $\epsilon>0$. The property of the supremum says that there exists is a $x_j$ such that $|sup_{k=1}^\infty{|x_k|}-\underbrace{max_{k=1}^{N}{|x_k|}}_{=x_j}|<\epsilon_1$. This means, there exists a $N \in \mathbb{N}$ such that $x_j=max_{k=1}^{N}|x_k|$. Choosing N such that the $\epsilon_1$ I get from the supremum is $\epsilon_1 \leq \epsilon$, then the inequality $|sup_{k=1}^\infty{|x_k|}-max_{k=1}^{N}{|x_k|}|<\epsilon$ holds. Now if $|x_{N+1}|>|x_j|$, the inequality still holds (Case 1).

If $|x_N+1|<|x_j|$ (Case 2), $ x_j=max_{k=1}^{N+1}{|x_k|}$ and the equality still holds.

And by induction it is true for all $n \geq N$.

Meaning, $\forall \epsilon >0 :\exists N \in \mathbb{N}$ such that $|sup_{k=1}^\infty{|x_k|}-max_{k=1}^{N}{|x_k|}|<\epsilon$ for all $n \geq N$

Answer 1

You are arguing a bit too complicated. Let $c=\sup_{k=1}^{\infty}|x_k|$. Given $\epsilon > 0$ there is an index $N$ such that $c - \epsilon < |x_N|$. Also $|x_j| \le c$ for all indices $j$. It follows that $$ c-\epsilon < \max_{k=1}^n|x_k| \le c $$ for $n \ge N$, and that proves the convergence.

Remark: You emphasize that this is a statement about a sequence of absolute values, but that is not really relevant. In the same way one can prove that if $(y_n)$ is any bounded sequence of real numbers then $$ \lim_{n \to \infty}\left( \max_{k=1}^n y_k \right) = \sup_{k=1}^{\infty} y_k \, . $$