How do I prove that $n^2$ diverges to infinity? I know that divergence implies that for every number $M$, there is an integer $N$ such that sn is less than or equal to $M$ whenever $n$ is greater than or equal to $N$. I also know that my proof begins with the inequality $n^2$ greater than or equal to M. where do I go from here?
 A: You got the definitions confused.
If a sequence diverges then it will always be greater than a certain arbitrary number $L$. If $u_n = n^2$ diverges, then for any $L$ that you choose I must find some $n$ such that $u_n$ is greater than $L$. We're essentially $racing$ between $L$ and $n$ to prove that $u_n$ diverges with the argument that I can always find a term for $u_n$ which is greater than any number you choose.
In mathematical notation, for any $n \in \mathbb{N}$:
$$\forall_{L \in \mathbb{N}} \quad \exists_{N \in \mathbb{N}} : n > p \Rightarrow u_n = n^{2} > L$$
$L$ is a natural number, hence it is positive. Choosing $n$ to be greater than $\sqrt{L}$ gives us $u_n$ to always be greater than $L$. For instance, let's say you choose $L = 4$. I then choose $n$ to be any number greater than $\sqrt{4}$, for example, $\sqrt{4} + 1$. In fact:
$$u_{\sqrt{4} + 1} = u_{3} = 3^2 = 9 > 4 = L$$
Let's say now you choose $L$ to be $100$. Well, then I choose any $n > \sqrt{100}$, for example, $n = \sqrt{100} + 1$. Then:
$$u_{11}=11^2 = 121 > 100 = L$$
As you can see, I'm able to always find $n$ such that $u_n$ is greater than any $L$ you choose. I just need to make $n > \sqrt{L}$.
A: You have the definition confused. For a sequence to diverge to infinity means that for every $M$ there is an $N$ such that $n \gt N $ implies that $s_n \gt M$
So you want to choose an $N$ as a function of $M$. i.e. choose $N = \sqrt{M}$ then $ n \gt N \implies s_n\gt M$. Since M is arbitrary we can make it as large as we want. Hence $s_n \to \infty$
