Proof explanation: "Every convergent sequence is bounded"

Question

I am looking at this proof:

Let $(a_n)_{n\in \mathbb{N}}$ be a sequence with $a_n \to a$. Choose $\varepsilon =1$, then there is a $N\in \mathbb{N}$ such that $|a_n-a|<\varepsilon$ for all $n\geq N$. $$|a_n|=|(a_n-a)+a|\leq |a_n-a|+|a|<1+|a| \quad \text{for all } n\geq N$$

Choose $A:=\max\{1+|a|, |a_1|, ..., |a_{N-1}|\}$. It follows that $|a_n|\leq A$ for all $n\in \mathbb{N}$

I don't understand what $A:=\max\{1+|a|, |a_1|, ..., |a_{N-1}|\}$ means or why one would choose $A$ like that. That's a problem I'm always facing with different proofs.

Answer 1

The proof argument here basically is

That if $\{a_n\}$ converges to limit a there will be an index $N_1$ from which upwards the sequence is bounded by the constant $1+|a|$

This is used then to imply the boundedness for all terms of the sequence $\{a_n\}$.

Since we know that there are only finitely many terms of the sequence (all $a_n$ with $n<N_1$) which possibly can be greater than the constant $1+|a|$, we have nothing more to do than to choose the maximum out of these numbers $a_1$ to $a_{N_1}$, $1+|a|$ and this will be an upper bound.

Answer 2

This is an outline of the proof to give you some orientation.

We have a convergent sequence. Then we cut it into two -- the initial part which has a finite number of terms, and the tail with infinitely many parts. Since it converges, we can choose the tail so that all terms in the tail are a distance of one or less from the limit. With this we can show that this tail is bounded by manipulating inequalities.

Finally, we consider the first (finite) part. Since it is a finite set of real numbers, it must have a minimum and maximum number. The maximum number is chosen as an upper bound for the head. Finally, we want to join the head to the tail. Then it makes sense (doesn't it?) to take the bigger of the two bounds (the one for the head and the other for the tail) as the upper bound for the whole sequence, and we are done.