# Prove: Convergent sequences are bounded

I don't understand this one part in the proof for convergent sequences are bounded.

Proof:

Let $s_n$ be a convergent sequence, and let $\lim s_n = s$. Then taking $\epsilon = 1$ we have:

$n > N \implies |s_n - s| < 1$

From the triangle inequality we see that: $n > N \implies|s_n| - |s| < 1 \iff |s_n| < |s| + 1$.

Define $M= \max\{|s|+1, |s_1|, |s_2|, ..., |s_N|\}$. Then we have $|s_n| \leq M$ for all $n \in N$.

I do not understand the defining $M$ part. Why not just take $|s| + 1$ as the bound, since for $n > N \implies |s_n| < |s| + 1$?

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$|s|+1$ is a bound for $a_n$ when $n > N$. We want a bound that applies to all $n \in \mathbb{N}$. To get this bound, we take the supremum of $|s|+1$ and all terms of $|a_n|$ when $n \le N$. Since the set we're taking the supremum of is finite, we're guaranteed to have a finite bound $M$.

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Taking the Max ensures that we take into account elements of $a_n$ for $n \leq N$ that could be bigger than $|s| + 1$? –  CodeKingPlusPlus Oct 15 '12 at 3:05
@CodeKingPlusPlus Yes, you got it right! –  Ayman Hourieh Oct 15 '12 at 8:36
Helpful even two years later! –  mathjacks Oct 15 '14 at 3:45

Because you want to be sure that the bound is large enough to ensure that $|s_n|\le M$ for all $n\in\Bbb N$, not just for all $n>N$. Taking $M\ge|s|+1$ ensures that the only possible exceptions to $|s_n|\le M$ are $s_1,\dots,s_N$, and taking $M\ge\max\{|s_1|,\dots,|s_N|\}$ takes care of these as well.

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Let's prove uniqueness first. Suppose the sequence has two limits, a and a'. Take any > 0. Then there is an integer N such that:

| aj - a | <


if j > N. Also, there is another integer N' such that

| aj - a' | <


if j > N'. Then, by the triangle inequality:

| a - a' | < | a - aj + aj - a' |
|aj - a | + | aj - a' |
< + = 2


if j > max{N,N'}. Hence | a - a' | < 2 for any > 0. But that implies that a = a', so that the limit is indeed unique.

Next, we prove boundedness. Since the sequence converges, we can take, for example, = 1. Then

| aj - a | < 1


if j > N. Fix that number N. We have that

| aj | | aj - a | + | a | < 1 + |a|


for all j > N. Define

M = max{|a1|, |a2|, ...., |aN|, (1 + |a|)}


Then | aj | < M for all j, i.e. the sequence is bounded as requi

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–  Shaun Oct 24 '14 at 9:23