Show that $\sqrt{n+1}-\sqrt{n}\to0$ Let $\ a_n=\sqrt{n+1}-\sqrt{n}$. I have to show that $\lim_{n\to \infty}a_{n}=0$.   
How should I start? Do I have to use any theorem?
 A: Use the fact that 
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}.$$
A: Hint: Think about the difference of squares formula $a^2-b^2=(a-b)(a+b)$.  You'd rather deal with $(\sqrt{n+1})^2-(\sqrt{n})^2$ than with $\sqrt{n+1}-\sqrt{n}$, right?
A: Use the mean value theorem for the function 
$f(x) =\sqrt{x}$ , $f'(x)=\frac{1}{2\sqrt{x}}$
Then $\sqrt{n+1}- \sqrt{n}= f(n+1)- f(n) = f'(\xi) \leq \frac{1}{2\sqrt{n}}$
A: Another solution based on applying a binomial expansion that converges for $n > 1$:
$$\begin{align} a_n &= \sqrt{n+1} - \sqrt{n} \\
& = \sqrt{n}\left(\sqrt{1 + \frac{1}{n}} - 1\right) \\
& = \sqrt{n}\left(\sum_{j=1}^\infty \frac{\Gamma(3/2)}{j! \Gamma(3/2 - j)} n^{-j}\right) \\
& \approx \frac{1}{2\sqrt{n}} - \mathcal{O}(n^{-3/2}) \\
& \rightarrow 0. \end{align}$$
A: I always felt like coming up with the binomial formula requires having seen that trick before. Below, I present a less elegant yet more straightforward approach.
If the claim were false, the numbers $ s:=\sqrt{n+1} $ and $ r:=\sqrt{r} $ would satisfy 
$$ s^2-r^2=1$$
 while having distance bounded below even for large $ n $. But for any $\epsilon> 0$ we have $ (r+\epsilon)^2=r^2+2\epsilon +\epsilon^2> r^2+2r\epsilon$. Inserting $ \epsilon:=s-r $ shows that actually $ r $ and $ s $ cannot be more than $1/(2r)$ away. The claim follows since $ r \to \infty $ as $ n\to \infty $. 
What is going on in this proof is: you wanna make a claim about a function which is defined to be the inverse of the square function. Therefore, you see what the claim would imply for the square function. Unlike the square root function, the square function is straightforward to work with.
