# Is there a proper theorem or proposition for compare limit of two sequence?

I am trying to use the fact that $\lim_{n \to \infty}\frac{1}{\sqrt{n}}=0$ to prove that if $a_n=\sqrt{n+1} - \sqrt{n}$ then its limit is $0$

The only thing that bothers me is what theorem or proposition should I use to state: If $b_n>a_n$, then $\lim_{n\to \infty}b_n>\lim_{n\to \infty}a_n$.

In this case $b_n$ is $\frac{1}{\sqrt{n}}$ and the $\lim_{n \to \infty}a_n=0$.

Can anyone please help on this question? (and perhaps how to edit format?)

Thanks in advance

• Also, if you want to learn how to format math for display on this site, look up how to use Latex. I think there's a link to an article somewhere on the Math StackExchange FAQ that can get you started. – Danny Mar 22 '15 at 5:17

## 2 Answers

hint: $0< a_n = \dfrac{1}{\sqrt{n+1}+\sqrt{n}} < \dfrac{1}{2}\cdot \dfrac{1}{\sqrt{n}}$

It may be easier to note that $\sqrt{n+1}-\sqrt{n} = { (\sqrt{n+1}-\sqrt{n}) (\sqrt{n+1}+\sqrt{n}) \over \sqrt{n+1}+\sqrt{n} } = {1 \over \sqrt{n+1}+\sqrt{n} }$.