Using the definition of convergence , show that $a_n=\sqrt{n+1}- \sqrt{n}$ Converges to $0$.

Question

How do I apply this finding to the definition:

So far I have got that:

\begin{align} a_n &=\sqrt{n+1}-\sqrt{n} \\ &=\left(\sqrt{n+1}-\sqrt n\right)\left(\frac{\sqrt{n+1}+ \sqrt n}{\sqrt{n+1}+ \sqrt n} \right)\\ &=\frac{n+1-n}{\sqrt{n+1}+ \sqrt n} \\ &=\frac1{\sqrt{n+1}+ \sqrt n} \end{align}

Answer 1

Since $$\sqrt{n+1}+\sqrt{n} \geq \sqrt{n} + \sqrt{n},$$

we have

$$0 \leq \lim_{n \to \infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} \le \lim_{n \to \infty} \frac{1}{2\sqrt{n}}=0$$

Answer 2

$|a_n|=\left|\dfrac{1}{\sqrt{n+1}+\sqrt n}\right|\lt$

$\left|\dfrac{1}{2\sqrt n}\right|.$

Let $\epsilon >0$ be given.

There is a $n_0 \in \mathbb{N}$ such that

$n_0 > \dfrac{1}{4\epsilon^2}$, i.e. $\epsilon > \dfrac {1}{2\sqrt{n_0}}$(Archimedes).

Then for $n\ge n_0 :$

$|a_n| \lt \dfrac {1}{2\sqrt n} \le \dfrac{1}{2\sqrt{n_0}} \lt \epsilon$.