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Prove $\displaystyle\lim_{n\to\infty} n=\infty$ using the limit definition for a converging sequence.

What I did: Suppose by contra position that $n$ tends to a finite real limit $L$, so from the definition of a converging sequence we have: $|n-L|<\epsilon\Rightarrow n<\epsilon+L$ so if we'll choose $N$ to be $N=\epsilon+L$ then because by definition the above is true for all $n>N$ we'll have: $\epsilon+L+1<\epsilon+L$ contradiction.

But apparently it's wrong and I was told I can't simply choose $N$, but why not? The definition states such $N$ exists...

Note: we use a definition with a real $N$.

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    $\begingroup$ This argument only shows the limit is not converging to a real number. The sequence $(1,-1,1,-1,...)$ also does not converge to any real number, and $\lim\limits_{k\to\infty} (-1)^k\neq\infty$. $\endgroup$ – Ian Mateus Dec 14 '14 at 13:34
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I suppose you mean contradiction, not contraposition.

As Ian Mateus points out in the comments, assuming it converges to a finite number is not the negation of diverging to infinity.

You can prove it directly. By definition, $$\lim_{n\to\infty}n=\infty \iff \forall M\in ]0,+\infty[\exists N\in \mathbb N\forall n\in \mathbb N(n\ge N\implies n>M).$$ So take $M$ as above, let $N$ be a natural number greater than $M$ and it follows easily.

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  • $\begingroup$ Wait so the answer by user38584 is wrong? $\endgroup$ – shinzou Dec 14 '14 at 13:39
  • $\begingroup$ @kuhaku I commit to this: none of the other answers at the moment proves that $\lim \limits_{n\to \infty}(n)=\infty$. $\endgroup$ – Git Gud Dec 14 '14 at 13:41
  • $\begingroup$ No kidding, they actually proved it in our lecture almost exactly as user38584 did... $\endgroup$ – shinzou Dec 14 '14 at 13:44
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    $\begingroup$ Git Gud is absolutely right, proving the result is easy, but certainly not acieved by showing there is no finite limit. $\endgroup$ – André Nicolas Dec 14 '14 at 13:49
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    $\begingroup$ @kuhaku Using just the definition of converging (to a real number) sequence, it's not possible. It misses cases like $\left(\sin \left(\frac{k\pi}2\right)\right)_{k\in \mathbb N}$. $\endgroup$ – Git Gud Dec 14 '14 at 14:04
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You can not simply choose an $N$ because that is not what the definition states. It says for any $\varepsilon>0$, there exists an $N\in\mathbb{N}$ such that $|x_n-L|<\varepsilon$ for all $n\ge N$. That means you have to choose an $\varepsilon$ and then find an $N$.

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  • $\begingroup$ Yes, I agree with this. I was a little confused with what the question was asking. $\endgroup$ – Math1000 Dec 14 '14 at 13:27
  • $\begingroup$ Aha, I see. Well thanks. $\endgroup$ – shinzou Dec 14 '14 at 13:27
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The basic question is how do you know that there exists at least an $n\in \mathbb{N}$ such that $n>N$? For this you need to prove the unboundedness of $\mathbb{N}$ which is what you are asked to prove. So basically your argument is circular and hence wrong.

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  • $\begingroup$ Can't I assume that $n>N$ is true from the limit definition? $\endgroup$ – shinzou Dec 14 '14 at 13:32
  • $\begingroup$ No. You can't. For this you have to prove that at least such an $n$ exists which would satisfy the inequality. $\endgroup$ – user170039 Dec 14 '14 at 13:34
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Suppose the sequence $x_n = n$, $n=1,2,3,\ldots$ has a limit $L$. Then for any $\varepsilon > 0$, there exists $N$ such that $n\geqslant N$ implies $|x_n-L|<\varepsilon$. So take $\varepsilon = 1$ and choose such an $N$. Let $m = 2 + \max\{N, \lceil L \rceil\}$ (where $\lceil\cdot\rceil$ denotes the ceiling function). Then $m\geqslant N$ but $m-L\geqslant2>1=\varepsilon$, a contradiction.

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