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


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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    $\begingroup$ Okay, this is may be a strange question, but wouldn't the triangle inequality say: $|s_{n}-s| \leq |s_{n}|+|s|$? $\endgroup$ – QuantumEyedea Oct 6 '15 at 1:12
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    $\begingroup$ Actually, no. Using the standard Triangle Inequality here is a tad naïve. I believe what Ross implies is $\vert s_n - s\vert \ge \vert s_n \vert - \vert s \vert < 1 \implies \vert s_n \vert < 1 + \vert s\vert$ $\endgroup$ – user203509 Feb 23 '16 at 23:54
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    $\begingroup$ We could still use Triangle Inequality here. Begin with $|s_{n}| = |s_{n}-s+s| \leq |s_{n}-s|+|s|$, which implies $|s_{n}| - |s| \leq |s_{n}-s|$. It follows that $|s_{n}| - |s| \leq |s_{n}-s| < 1$. $\endgroup$ – nexolute Mar 1 '16 at 6:13
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    $\begingroup$ @Greg.Paul I doubt you still care and you probably know this already, but for others, the result $|s_n|-|s|\leq|s_n-s|$ follows as in nexolute's comment above and is called the Reverse Triangle Inequality. It is usually invoked without proof as it is considered well-known. $\endgroup$ – The Count Jan 2 '17 at 16:04

$|s|+1$ 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 $M$.

  • $\begingroup$ Taking the Max ensures that we take into account elements of $a_n$ for $n \leq N$ that could be bigger than $|s| + 1$? $\endgroup$ – CodeKingPlusPlus Oct 15 '12 at 3:05
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    $\begingroup$ @CodeKingPlusPlus Yes, you got it right! $\endgroup$ – Ayman Hourieh Oct 15 '12 at 8:36
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    $\begingroup$ Helpful even two years later! $\endgroup$ – mathjacks Oct 15 '14 at 3:45
  • $\begingroup$ Instead of triangle inequality, can we do it like this ? $ s-1<s_n<s+1 $ whenever $n>N$. Now take lower bound of sequence to be $\text{min}\{s_1,s_2,...,s_N,s-1\}$ and upper bound of sequence to be $\text{max}\{s_1,s_2,...,s_N,s+1\}$. $\endgroup$ – MeetR Aug 16 '19 at 14:36

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.

  • $\begingroup$ Do you mean the following : as |$S_{n}-S$|<$1$ holds only when $n>N$ so is |$S_{n}$|<|$S$|+$1$. Which implies that there might be another number $M$ such that M>|S|+1 s.t. it is the bounded. So you are assuming for some $|S_{k}|$ where k is less than equal to $N$ , we can not guarantee that the current upperbound $|S|+1$ holds so you take the max of all the such $|S_{k}|$ and again there is chance between that max{$|S_{k}|$} and the $|S|+1$ so again take the max and name that $M$, NOW MY QUESTION is how you can guarantee |$S_{n}$|<= M for all n, why not strict inequality? $\endgroup$ – NewBornMATH Feb 23 '19 at 9:40
  • $\begingroup$ @NewBornMATH because M is Sn $\endgroup$ – Milan Jun 9 '19 at 16:51

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