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I need to prove that $(n+1)^{1/2} - n^{1/2}$ is null. I know there's a similar question already posted but I need to prove this stating rules etc.

What I've done already is use difference of 2 squares to get to:

$1/[(n+1)^{1/2} + n^{1/2}]$ which I'm sure is near the end. Do I need to use the Squeeze Rule?

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"Squeeze Rule"? This sounds a bit gruesome ;) –  Xabier Domínguez Apr 2 '12 at 17:28
    
There's a result that states that if $a_n$ is a sequence with $a_n = \frac{1}{b_n}$ for each $n$ then $a_n \to 0$ if and only if $\left| b_n \right| \to \infty$. It's easy to prove and is very useful here. –  Clive Newstead Apr 2 '12 at 17:28
    
You are not using squeeze theorem here. –  sdcvvc Apr 2 '12 at 17:29
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1 Answer 1

up vote 2 down vote accepted

We have $a_n=\sqrt{n+1}-\sqrt{n}$, and we want to show that $\lim a_n=0$.

Let see, what is $\sqrt{n+1}-\sqrt{n}$:

$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$

So, when $n\to\infty$, we get $\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$.

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