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Convergent sequence in $\Bbb R$ is bounded. Proof:

In the definition of a convergent sequence:

$$(\forall \varepsilon>0), (\exists n_\varepsilon\in\Bbb N), (\forall n\in\Bbb N), ((n>n_\varepsilon)\Rightarrow(|a_n-a|<\varepsilon))$$ let $\varepsilon=1$, then there exists $n_\varepsilon\in\Bbb N$ such that $(n>n_\varepsilon)\Rightarrow(|a_n-a|<1$). Now for $n>n_\varepsilon$ we have $|a_n|\leq|a_n-a|+|a|\leq 1+|a|$. Let $M=\max\{|a_1|,...,|a_{n_\varepsilon}|,1+|a|\}$ Then $\forall n\in\Bbb > N, \ |a_n|\leq M$, i.e. the sequence is bounded.

What's the idea behind this proof? I understand the first part, but then when they define $M$ I'm lost. I don't see how is that related to the first part of the proof.

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    $\begingroup$ @ΘΣΦGenSan No problem, I always tend to upvote all the answers I get to show that I appreciate them. $\endgroup$ – lmc Feb 6 '17 at 12:59
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The idea is that after a while (i.e., after $n_\epsilon$), the sequence is bounded by $1+|a|$, and before that, the sequence is finite so it is bounded by the largest element.

What you do is you split the sequence into two parts, a starting part and "the rest"

  1. The starting part, $a_1,a_2,\dots, a_{n_\epsilon}$, is clearly bounded by $M_1=\max\{a_1,a_2,\dots, a_{n_\epsilon}\}$. It's bounded by $M_1$ because $M_1$ is the largest element of a finite set, and so it must be larger than all elements of that set.
  2. The rest of the sequence, $a_{n_\epsilon+1}, a_{n_\epsilon+2},\dots$, is infinite, so we can't just take its max. However, we know and can show that this part must be bounded by $M_2=1+|a|$.

Now, the first part of the sequence is bounded by $M_1$, and the second by $M_2$, so both together must be bounded by $\max\{M_1,M_2\}$, since $M\geq M_1, M_2$.


For example, you need to define $M$ because $1+|a|$ may not be the upper limit of the entire sequence. For example, the sequence $\frac2n$ has a limit of $0$, but the sequence is not bounded by $1+|0|$ since $2$ is also a part of the sequence.

However, the sequence is bounded by $\max\{2, 0+1\}=2$.

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    $\begingroup$ Thanks so much, I wouldn't have ever figured it out by myself. $\endgroup$ – lmc Feb 6 '17 at 13:04
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    $\begingroup$ @Lewis You're very welcome. The method of splitting the sequence into a finite and convergent part is actually quite common, as intuitively, anything that's finite shouldn't cause too much problems, so you start to only think about $a_n$ for large values of $n$, and then mop up the first couple of values (even if, of course, that might mean billions of values -- if it's finite, it's fine). $\endgroup$ – 5xum Feb 6 '17 at 13:08
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The idea behind this proof is to separate the values of the sequence into 2 parts : the ones that are close to the limit (within distance 1 of it) and the ones that are "far" from the limit.

By definition of convergent sequences, those last ones are in finite number, so the set of their absolute values is bounded, by a real $M$. And the set of the others is bounded by $|l|+1$.

So the whole set of absolute values is bounded by $\max(|l|+1,M)$.

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See the main idea is that ,, by the convergence property , you get a bound of an infinite part of the sequence i.e of all the $x_n$'s after some stage. Now then the bound becomes easy to define , just take the maximum of all the $x_n$'s before that particular $N_0$ with the bound you have got for the infinite part. Hence the bound of the sequence gets defined.

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We want some $M>0$ so that $|a_n|\leq M$ for all $n\in\Bbb N$ and if this happens, we say that the sequence $(a_n)$ is bounded. Now, you already mentioned that $$|a_n|<1+|a|\quad\text{for }n=n_{\epsilon}+1,n_{\epsilon}+2,n_{\epsilon}+3,\dots.\tag{$*$}$$ What we gonna do to the numbers $|a_1|,|a_2|,|a_3|\dots,|a_{n_{\epsilon}}|$? This give an idea of taking $$M=\max\{|a_1|,...,|a_{n_\varepsilon}|,1+|a|\}.$$ So, using the definition of our $M$ together with $(*)$, we get $$\begin{align} &|a_1|\leq M\\ &|a_2|\leq M\\ &|a_3|\leq M\\ &\quad\vdots\\ &|a_{n_{\epsilon}}|\leq M\\ &|a_{n_{\epsilon}+1}|< 1+|a|\leq M\\ &|a_{n_{\epsilon}+2}|< 1+|a|\leq M\\ &|a_{n_{\epsilon}+3}|< 1+|a|\leq M\\ &\quad\vdots\\ \end{align}$$ With the above, we conclude that $|a_n|\leq M$ for all $n\in\Bbb N$.

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