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I was seeing Prove: Convergent sequences are bounded. The proof is as follows:

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've some problem in understanding the approach of this deduction.

We need to show that the sequence is bounded which means $|s_n| \le M$ . Now, this must be for every $n$. What I am not understanding is whether the first step to show that for $n \gt N$, $|s_n| \lt |s| +1$ was necessary. After all, a number $M$ greater than all the sequence-elements were taken next; so why not, take this step firstly?

I only want to know how the first step for showing that for $n \gt N$, $|s_n| \lt |s| +1$ is essential for the proof. Also, it deduced that $|s_n| \color{red}{\lt} |s| +1$; so how did, finally, at the last step of proof $\color{\red}\le$ come in place of $\color{red}{\lt}$? Thanks in advance.

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$M$ was not chosen to be greater than all terms of the sequence: it was chosen to be at least as large as $|s|+1$ and the first $N$ terms of the sequence.

The first step takes care of all but the first $N$ terms: there is some integer $N$ such that $|s_n|<|s|+1$ whenever $n>N$. Thus, $|s|+1$ is almost an upper bound for the sequence: it’s bigger than the absolute value of every term except possibly one or more of the terms $s_1,s_2,\ldots,s_N$, However, there are only finitely many of those terms, so we can pick the one with largest absolute value; say that it’s $s_m$. Now we know that $|s_k|\le |s_m|$ if $1\le k\le N$, and $|s_k|<|s|+1$ if $k>N$. If we now let $M=\max\{|s|+1,|s_m|\}$, we know that

$$|s_k|\le|s_M|\le M$$

if $1\le k\le N$, and

$$|s_k|<|s|+1\le M$$

if $k>N$, so $|s_k|\le M$ for all $k$. Thus, the sequence lies entirely in the interval $[-M,M]$ and is therefore bounded.

As for your last question, if $a<b$, then it’s certainly also true that $a\le b$: if you know the former inequality, you can certainly claim the latter.

The basic idea of the proof is that since the sequence is convergent, it must eventually get and stay close to its limit $s$. In this specific argument we choose $N$ so that $|s_n-s|<1$ whenever $n>N$, so that the sequence stays within one unit of $s$. The first $N$ terms could be anywhere, but there are only finitely many of them, so there is a largest (in absolute value). Thus, we have the bound $|s|+1$ that is known to work for all but the first $N$ terms, and we can get a bound on that finite set of terms. We then take the larger of these bounds and can be confident that it bounds every term.

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  • $\begingroup$ Sir, thanks for the answer. Shouldn't in $s_k\le|s_M|\le M$ & $s_k<|s|+1\le M$, there would be $|\;\;|$ in $s_k$; in the above lines you've written them correctly including $|\;\;|$ sign: "Now we know that $|s_k|≤|s_m|$ if $1≤k≤N$, and $|s_k|<|s|+1$ if $k \gt N$"; but later, you didn't mention them. So, I think this is a typo:| $\endgroup$
    – user142971
    Aug 19 '15 at 3:59
  • $\begingroup$ Also, I want to know how you inferred $|s_k|\le M$ from $|s_k|<|s|+1\le M$; Yes it is accurate to write that $|s_k| \lt M$ but why the equality? $|s|+1$ can be equal to $M$ but not $|s_k|$ since it is less than $|s|+1$. So, can you help me shun this confusion? $\endgroup$
    – user142971
    Aug 19 '15 at 4:08
  • $\begingroup$ @user36790: Yes, I inadvertently omitted the absolute value signs around $s_k$ in the displayed formulas; they’re fixed now. Thanks for noticing. \\ From $|s_k|<|s|+1\le M$ we can, as you say, actually infer that $|s_k|<M$, but of course if this is true, then it’s certainly also true that $|s_k|\le M$. For $1\le k\le N$, however, we can infer only that $|s_k|\le M$. Thus, the best that we can claim for all $k\in\Bbb Z^+$ is that $|s_k|\le M$. If we defined $$M=\max\{|s|+1,|s_1|+1,\ldots,|s_N|+1\}\;,$$ then we could infer that $|s_k|<M$ for all $k$, since for $1\le k\le N$ we would have ... $\endgroup$ Aug 19 '15 at 4:15
  • $\begingroup$ ... $|s_k|<|s_k|+1\le M$. But we don’t need this: in order to show that that sequence is bounded, we need only show that there is an $M$ such that $|s_k|\le M$ for all $k$, and we’ve done that. (It may help to point out that it’s perfectly correct to write, for instance, $2\le 3$, even though we know that in fact the stronger statement $2<3$ is true.) $\endgroup$ Aug 19 '15 at 4:16
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The point is that the definition of convergence only says that given any positive error margin eventually the elements in the sequence will be within that error margin around the limit. We do not know when that eventual situation would be, so we have no choice but to separately bound the initial part of the sequence before that point. Thankfully that initial part has finitely many numbers and hence is bounded, so since the latter part is also bounded, the whole thing is bounded.

Also, if $a < b$ then $a \le b$ so there is nothing wrong with doing that last step.

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According to definition of convergent sequence after certain stage all terms of sequence lies in a band. I.e $L-\epsilon<a_n<L +\epsilon$ after n greater than equal to m say. This means after m, the sequence is bounded above by $L+\epsilon$. So we need to worry about terms from $n=1$ to $n=m-1$. Let we have P be the maximum value of these terms and let us take $M=\max\{P,L+\epsilon\}$. So sequence is bounded above by $M$.

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  • $\begingroup$ This is a very common proof technique. $\endgroup$ Aug 19 '15 at 5:19

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