A difficulty in understanding the proof of "Every convergent sequence is bounded"

Question

The proof is given below, but I could not understand:

why he considered exactly those $N+1$ positive numbers?

why if $n \geq N$ there are those two possibilities ?

could anyone explain them for me?

Answer 1

For the first question, it is because you have complete control over what happens at $N$ and beyond: everything is trapped within $1$ of $L$. What you don't know is the relation of the terms $a_1, a_2, \ldots, a_{N-1}$ to $L$. But they're finite in number, so we're fine.

Secondly, there are two possibilities because you don't know ahead of time if a term $a_k$ is negative or non-negative. Two quick arguments settle it all.

The key is to note that $M$ is chosen to be at least as big $|L+1|$, $|L-1|$, and any of the previous terms. I actually think the argument should have stopped at the sentence "Then $|a_n| \leq M$ for all $n \in \mathbb{N}$" because it's (a) clear that you're done and (b) the rest of the paragraph makes it seem harder than it is.

Answer 2

Since, for $n\ge N$ we know that $|a_n|< max\{|L-1|,|L+1|\}$ and we don't know whether anyone of $\{|a_1|,|a_2|,..|a_{N-1}|\}$ is greater than $max\{|L-1|,|L+1|\}$ or not. So,we considere the maximum of all this numbers in order to find an upper bound for $|a_n|$

Answer 3

The statement is that if $n\ge N$ there are two possibilities, which are $a_n \ge 0$ or $a_n <0$ and you are asking why these two possibilities are the result of $n\ge N$.

Your confusion is justified because whether or not $n\ge N$, the only two possibilities are $a_n \ge 0$ or $a_n <0$.

What the author is trying to say is that the case of $n< N$ has already been discussed so now we concentrate on the case of $n\ge N$

Answer 4

Convergence definition demands something happen for every positive number $\epsilon$. Let us take (my favourite positive number) $\epsilon= \frac47$. If the limit of the convergent sequence is $L$, then all the terms from and beyond specific one (say 883rd term) lie in between $L-\frac47$ and $L+\frac47$.

Let the biggest and the smallest of the the first 882 terms of the sequence be called $B$ and $s$.

Now among the 4 numbers, $B, s, L+\frac47, L-\frac47$ call the biggest number BIG and the smallest number TINY.

Now all the terms of the sequence (without exception) lie in between TINY and BIG, and hence the convergent sequence is bounded.

Answer 5

An example. Consider the converging sequence $a_n=\frac{3n}{n+1}$ with $\lim_\limits{n\to\infty} \frac{3n}{n+1}=3$. We will show that the sequence ${a_n}$ is bounded. Let $\epsilon=1$ and choose $N$ such that if $n\ge N$, $|a_n-3|<\epsilon=1$. The inequality is equivalent to $$-1<a_n-3<1 \iff 2<a_n<4.$$ Now let's calculate a few terms: $$a_1=1.5, a_2=2,a_3=2.25, a_4=2.4, a_5=2.5, ...$$ Note that for $n\ge 3$, it holds true: $$2<a_n<4 \iff 2<\frac{3n}{n+1}<4 \iff 2<n.$$ Note that $(2,4)$ is the bound for $a_n, n\ge 3$, but we need to find bounds for all $n\ge 1$.

Now consider: $$|a_1|=1.5, |a_2|=2, |3-1|=2, |3+1|=4.$$ The maximum of them is $4$, hence: $$|a_n|<4 \iff -4<a_n<4, n\ge 1.$$ You can also state: $$1.5\le a_n<4, n\ge 1.$$