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A general proof of the theorem that "convergent sequences are bounded" is that:

$$\forall \varepsilon>0, \exists N \in\Bbb N, \forall n\in\Bbb N, n≥N\Rightarrow|a_n-L|<\varepsilon$$ let $\varepsilon=1$, then there exists $N\in\Bbb N$ such that $(n≥N)\Rightarrow(|a_n-L|<1$). Thus, for $n≥N,$ we have $|a_n|\leq|a_n-L|+|L|\leq 1+|L|$ by the triangle inequality. Let $M=\max\{|a_1|,...,|a_{N}|,1+|L|\}.$ Then $\forall n\in\Bbb > N, \ |a_n|\leq M$. Hence, the sequence is bounded.

I'm having trouble understanding this proof because I cannot seem to find the difference between $n$ and $N$. Because of this confusion, I am also confused on how we know that the set $n<N$ is finite.

Any verification/explanation would be great!

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    $\begingroup$ $N$ is a fixed number is found by convergence. $\endgroup$ – Nosrati Mar 30 '17 at 2:39
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    $\begingroup$ $n$ varies by sequences as progress. $\endgroup$ – Nosrati Mar 30 '17 at 2:42
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    $\begingroup$ Let $\Big\{\dfrac{5}{n}\Big\}_{n=1}^\infty$ is your sequence. you see $$\dfrac{5}{1},\dfrac{5}{2},\dfrac{5}{3},\dfrac{5}{4},\dfrac{5}{5},\dfrac{5}{6},\dfrac{5}{7},\dfrac{5}{8},\dfrac{5}{9},\cdots,\dfrac{5}{n},\cdots$$ $\endgroup$ – Nosrati Mar 30 '17 at 2:43
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    $\begingroup$ Here $L=0$ so for what $n$'s $|a_n|\leq 1+|L|=1$.? $\endgroup$ – Nosrati Mar 30 '17 at 2:46
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    $\begingroup$ The sequence converges. As you see above, $N=4$ and the rest terms are less than or equal $1$ then $M=\max\{|a_1|,...,|a_{N}|,1+|L|\}=\max\{\dfrac{5}{1},\dfrac{5}{2},\dfrac{5}{3},\dfrac{5}{4},1\}=5$ $\endgroup$ – Nosrati Mar 30 '17 at 2:51
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$n$ and $N$ are their own quantities; we could have as easily used $x,y, \Lambda$ or $i$ instead of $N$, but the author chose to use $N$ to signify that there is some relation between $n$ and $N$ (a good practice to adopt for your own proofs). In this definition,

  • $n$ indicates the index of our sequence: $a_1, a_2, \dots, a_n, \dots$
  • $N$ with $n \geq N$ indicates how far out in the sequence we need to go before some property holds; it might not be true that $|a_n - L|< \epsilon$ for the first ten/thousand/million/billion/... terms in our sequence, but eventually it always holds. $N$ establishes a sufficient distance down our sequence for this property to hold.

To see this in action, let's prove that $a_n = 1/n$ converges to $0$. Our sequence is $(1, 1/2, 1/3, 1/4, \dots)$.

Suppose we were given $\epsilon = 1/10$. It would not be true that $|a_n - 0| < 1/10$ for all of the terms of out sequence; in particular $|a_1 - 0| = |1-0| = 1 \not < 1/10$. However, this is eventually true! As such, we need to "throw away" some of our starting terms. If we look at our sequence starting from the 11th term, we have $(a_{11}, a_{12}, a_{13}, \dots) = (1/11, 1/12, 1/13, ...)$ and all of these terms satisfy $|a_n - 0| < 1/10$. As such, we would need to take $N\geq 11$. Note that $N=11$ is the smallest $N$ we could chose, but $N=1,000,347$ would also work.

However, to show convergence, we must be able to produce an $N$ when we are given any tolerance $\epsilon >0$. Our proof would go as follows:

Proof: Let $\epsilon >0$ be given. Take any integer $N > 1/ \epsilon$. If $n \geq N$, then $0<a_n = \frac{1}{n} \leq \frac{1}{N} < \frac{1}{1/\epsilon} = \epsilon $ as desired. $\square$

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    $\begingroup$ Thank you. This really helped me better understand the proof and convergence in general! $\endgroup$ – Mathgirl Mar 30 '17 at 3:07

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