limit of a radical function

Question

I want to show that $\lim_{n\to\infty}a_{n} = 0$ where $a_{n}=\sqrt{n+1}-\sqrt{n}$ $$ a_{n}={\sqrt{n+1}-\sqrt{n}\over1}\bullet{\sqrt{n+1}+\sqrt{n}\over\sqrt{n+1}+\sqrt{n}} \\ a_{n}={1\over\sqrt{n+1}+\sqrt{n}} $$ Now I know the denominator gets infinitly large so it will go to zero but I cant seem to show that it does go to infinity. Maybe... $$ {1\over\sqrt{n+1}+\sqrt{n}}>{1\over\sqrt{n}} $$ That root is confusing me, thank you.

Answer 1

the correct inequality is: $$\frac {1}{\sqrt{n+1}+\sqrt{n}}<\frac {1}{\sqrt{n}}$$ So, as $n\to\infty$ we have $\frac {1}{\sqrt{n}}\to0$, or $\frac {1}{\sqrt{n+1}+\sqrt{n}}\to0$

Answer 2

If you can prove that for any $U \in \Bbb{R}$ there is an $n \in \Bbb{R}$ such that $ \sqrt{n+1} + \sqrt{n} > U$ then your job is done.

The reason being $ \sqrt{n+1}+ \sqrt{n} $ is a strictly increasing function, so if you find a single n that's big enough every value after that will also be bigger. Furthermore by showing there isn't A single real number U that bounds the thing, and it's increasing, we conclude it tends to infinity.