# Does my proof about convergent sequences look ok: $\lim_{n\to \infty}\frac{\sqrt{n}}{n+1} = 0$?

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?

Yes, your proof is correct. ${ { { { } } } }$