Question about proving $\displaystyle\lim_{n\to\infty} n=\infty$ using the limit definition for a converging sequence 
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 $N$.
 A: 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 $N$.
A: 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.
A: 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.
A: 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.
