Prove that $$\lim_{n\to \infty}\frac{\sqrt{n}}{n+1} = 0$$
My proof: Let $\epsilon > 0$ be given. Let $N>\frac{1}{\epsilon^2}$. Then for all $n \geq N$, we have
$$\Bigg|\frac{\sqrt{n}}{n+1}-0\Bigg| = \frac{\sqrt{n}}{n+1} < \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{N}} < \epsilon$$
Does my proof about convergent sequences look ok?