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?
Solution 1:
Use the fact that
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}.$$
Solution 2:
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?
Solution 3:
Use the mean value theorem for the function $f(x) =\sqrt{x}$ , $f'(x)=\frac{1}{2\sqrt{x}}$
Then $\sqrt{n+1}- \sqrt{n}= f(n+1)- f(n) = f'(\xi) \leq \frac{1}{2\sqrt{n}}$