How to prove that a sequence diverges using the Cauchy definition? It was a question of my integral calculus exam:
Write the Cauchy definition of
$\lim_{n\rightarrow\infty}  a_n  = L$
So, I literally wrote:
" $\{a_n\}$ is a Cauchy sequence if given $\epsilon>0$ there exists $N \in \mathbb{N}$ such that $\forall\ m,n\in\mathbb{N}$ if $n,m\ge N$, then $|a_n - a_m|<\epsilon$, or with symbols:
$\forall\epsilon>0:\exists N\in\mathbb{N}:\forall\ m,n \in\mathbb{N}: m,n\ge N\implies |a_n - a_m|<\epsilon$.
And if $\{a_n\}$ is a Cauchy sequence, then $\lim_{n\rightarrow\infty}  a_n  = L$  ( i.e. $\{a_n\}$ converges) "
The following question was:
With this definition, find $\lim_{n\rightarrow\infty}  (3n+2)$
And that is what I did:
I proposed that $\lim_{n\rightarrow\infty}  (3n+2)   = \infty$ i.e.  $\{3n+2\}$ diverges.  So I think that I have to prove that this limit is equal to infinity using the Cauchy definition.  
But if a sequence $\{a_n\}$ diverges $\implies$ $\{a_n\}$ is not a Cauchy sequence.
And $\{a_n\}$ is not a Cauchy sequence if
$\exists\epsilon>0:\forall N\in\mathbb{N}:\exists\ m,n \in\mathbb{N}:$   $m,n\ge N$ and $|a_n - a_m|\ge\epsilon$ $(1)$
So, I must to prove $(1)$.
$(1)$ says that there exists $m,n\in\mathbb{N}$ such that ..., so $(1)$ has not to apply $\forall m,n\in\mathbb{N}$, and for this fact (I think) I can assign convinient values to $m,n$ in order to prove $(1)$:
If i choose $m=N+1,n=N$ then $|3(N+1)+2 -(3N+2)|=3>2$.
So exists $\epsilon=2>0 $ such that $\forall N \in\mathbb{N}:$
there exists $m=N+1, n=N$ such that $N+1,N \ge N$ and $|3(N+1)+2 -(3N+2)|=3>\epsilon$.
Hence $\{3n+2\}$ is not a Cauchy sequence $\implies$  $\lim_{n\rightarrow\infty}  (3n+2)   = \infty$.
Is my proof correct? If it is not correct, could you help me proving that, please?
 A: The definition of a Cauchy sequence is:
$\forall \epsilon > 0, \exists N \in  \mathbb{N}: \forall m,n \in \mathbb{N}: m,n \ge N \implies |a_m-a_n| < \epsilon$, and the negation of this definition is:
$\exists \epsilon > 0, \forall N \in \mathbb{N}, \exists m,n \in \mathbb{N}: m,n \ge N, |a_m - a_n| \ge \epsilon$.
For your sequence $a_n = 3n+2, n \ge 1$, choose $\epsilon = 1$ (your choice is $3$), for any $N \in \mathbb{N}$, let $n =N, m = N+1$. Observe $m, n \ge N$, and $|a_m - a_n| = |a_{N+1} - a_{N}|= |3(N+1) - 2 - 3N+2|= 3 > 1= \epsilon $. Thus the sequence above is not Cauchy. 
A: You did not write the def'n of $\lim_{n\to \infty}a_n=L.$ What you wrote in response to "Define what it means for $(a_n)_n$ to converge to $L$" was: " $(a_n)_n$ is a Cauchy sequence (which means ...) which converges to $L$."
The def'n of $\lim_{n\to \infty}a_n=L$ is $$\forall r>0\;\exists m\;\forall n>m\;(|L-a_n|<r).$$ I have omitted specifying that $m,n\in \mathbb N.$ It is conventional that in "$\lim_{n\to \infty}a_n$", the values of $n$ are restricted to members of $\mathbb N$ unless  stated otherwise. 
