Proof that a sequence is divergent Is my proof valid?
Suppose $a_n=\frac{1}{n^2}-\sqrt{n}$.
I will now proof that $a_n$ diverges:
$\forall \epsilon > o \exists n^*  \forall n \ge n^*: |a_n - L| < \epsilon$
So $|\frac{1}{n^2}|-|\sqrt{n}|-|L|<|\frac{1}{n^2}-\sqrt{n}-L|<\epsilon$
$\frac{1}{n^2}-\sqrt{n}-|L|<\epsilon$
$\frac{1}{n^2}-\sqrt{n}<\epsilon + |L|$
Suppose $z=\sqrt{n}$
$\frac{1}{z^4}-z=\frac{1-z^5}{z^4}<\epsilon+|L|$
$-\frac{z^5}{z^4}<\frac{1-z^5}{z^4}<\epsilon+|L|$
$-z<\epsilon+|L|$
$z>-(\epsilon+|L|)$
$n>(-(\epsilon+|L|))^2=(\epsilon+|L|)^2$
$\epsilon<\sqrt{n}-|L|$ but this is not valid for every $\epsilon$, so contraction and $a_n$ can't be converging for that reason and must be diverging.
 A: No, it’s not valid: no matter what $\epsilon$ and $L$ are, the inequality $\epsilon < \sqrt{n}-\vert L\vert$ is true for all sufficiently large $n$. Going back up the argument, you can’t possibly get a contradiction from $$\frac1{n^2}-\sqrt{n} < \epsilon + \vert L\vert:$$ the left-hand side is never positive for $n\ge 1$, while the right-hand side is always positive, so the inequality is true whenever $n\ge 1$.
Think about why the sequence diverges: $\dfrac1{n^2}$ approaches $0$, and $\sqrt{n}$ increases without bound, so $a_n$ must tend to $-\infty$. In other words, for any $L$ there must be some positive integer $n_L$ such that $a_n < L$ whenever $n \ge n_L$. The trick is to prove this rigorously.
Your idea of using a proof by contradiction is fine, but you should say what you’re doing. In your argument your third line should have been something like this: ‘Assume to get a contradiction that the sequence converges’. In the argument that I’m suggesting here, the first step is:

Suppose that there is some real number $L$ such that $a_n \ge L$ for all $n$, i.e., such that $\dfrac1{n^2}-\sqrt{n} \ge L$ for all $n$.

What does that imply that might be useful? We know pretty much what $\dfrac1{n^2}$ is doing, so let’s put it on the same side of the inequality as the fixed quantity $L$:

Then $\sqrt{n} \le \dfrac1{n^2}-L$ for all $n$.

Now $\dfrac1{n^2}-L \le 1+\vert L\vert$ (why?), so what happens when $n > (1+\vert L\vert)^2$?
