sequence convergent and divergent proof $\{ n^3 \}$ $\{ n^3 \}$ I know this approaches infinity therefore it diverges. How do I prove a sequence diverges? 
I guess I'll assume the contradiction that it converges.
$\exists L \in \mathbb R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $\left| n^3 - L \right| < \epsilon$
Let $\epsilon > 0$ be arbitrary
Choose $N = $ ____ > 0
Suppose $ n > N$, then .... I don't know from here
 A: To prove that a sequence diverges to infinity, one approach is to use the following:
A sequence ${a_n}$ approaches infinity if given any $M \in \mathbb{N}$, there exists $N \in \mathbb{N}$ such that for all $ n \geq N$ we have $a_n > M$.  
In this case given any $M \in \mathbb{N}$, if $N = M$, then for all $n \geq M = N$, we have $n^3 \geq M$. 
A: The approach you use will prove divergence (that is, non-convergence).  If you assume convergence to $L$ and arrive at a contradiction, you know convergence to $L$ is impossible. Ergo, non-convergence.
But that is not the same as saying the sequence diverges infinitely.  For that, you have to take a different approach.
Pick any arbitrary $M\in\mathbb{R}$ such that $M>0$.  Thus, $\forall \varepsilon>0$, and we must choose $N\in\mathbb{N}$. In particular let us choose $N> \sqrt[3]{M}$. This is is permissible by the Archimedean principle. And now $\forall n\ge N$,
We have $|n^3|=n^3\ge N^3 > (\sqrt[3]{M})^3 = M$. Thus, $|n^3|>M$.
All this means in plain English is that the sequence exceeds any arbitrary $M>0$ we choose if we pick a sufficiently large $N$ in the sequence.

Alternatively, it is also reasonable to show that $\lim\limits_{n\to\infty}\frac{1}{a_n}=0$.  This is logically equivalent to $\lim\limits_{n\to\infty}{a_n}=+\infty$
