Q:

Proving that this sequence converges using the epsilon definition

I have this sequence $\sqrt{n+1}-\sqrt{n}$ and I'm trying to get a better understanding how to apply the epsilon definition of convergence.
I want to prove that this sequence converges to $0$ using this definition.
Here are my steps so far:
$\forall\epsilon>0\exists N_0\in\mathbb{N}\forall n\geq n_0:|\sqrt{n+1}-\sqrt{n}|<\epsilon$

$|\sqrt{n+1}-\sqrt{n}|<\epsilon$
$\iff (\sqrt{n+1}-\sqrt{n})^2<\epsilon ^2$
$\iff (n+1)-2\sqrt{n}\sqrt{n+1}+n<\epsilon^2$
$\iff 2n+1-2\sqrt{n}\sqrt{n+1}<\epsilon^2$
$\iff 2n-2\sqrt{n}\sqrt{n+1}<\epsilon^2-1$
$\iff 2(n-\sqrt{n}\sqrt{n+1})<\epsilon ^2 -1$
$\iff n-\sqrt{n}\sqrt{n+1}<\frac{\epsilon ^2 -1}{2}$
From here on, I'm not completely sure if I'm doing it correctly
$\Rightarrow 2n-n^2=n-n^2+n=n-n(n+1)<n-\sqrt{n}\sqrt{n+1}<\frac{\epsilon^2 -1}{2}$
$\Rightarrow n^2> \frac{\epsilon^2 -1}{6}$
And now I don't know how to remove the exponent. I can't use the square root since the right hand side is not always positive.
What is the next step?

A:

Hint
Use that $$|\sqrt{n+1}-\sqrt{n}|<\epsilon\iff (\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})<\epsilon(\sqrt{n+1}+\sqrt{n}).$$ That is:
$$1<\epsilon(\sqrt{n+1}+\sqrt{n}).$$ 
Note that if $n\ge \dfrac{1}{\epsilon^2}$ the inequality is satisfied.

