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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?

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Use the fact that

$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}.$$

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    $\begingroup$ JOEF, Old John multiplied top and bottom by the conjugate, which is the two radical terms with a + inbetween. $\endgroup$ – imranfat Oct 30 '13 at 15:54
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    $\begingroup$ Yes, that is how it is proved. Then use the fact that the right-hand side clearly tends to zero. $\endgroup$ – Old John Oct 30 '13 at 15:55
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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?

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Use the mean value theorem for the function $f(x) =\sqrt{x}$ , $f'(x)=\frac{1}{2\sqrt{x}}$

Then $\sqrt{n+1}- \sqrt{n}= f(n+1)- f(n) = f'(\xi) \leq \frac{1}{2\sqrt{n}}$

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  • $\begingroup$ I have not done mean value theorem yet $\endgroup$ – JOEF Oct 30 '13 at 16:07
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    $\begingroup$ @joef I dont know what to say $\endgroup$ – clark Oct 30 '13 at 16:09
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Another solution based on applying a binomial expansion that converges for $n > 1$: $$\begin{align} a_n &= \sqrt{n+1} - \sqrt{n} \\ & = \sqrt{n}\left(\sqrt{1 + \frac{1}{n}} - 1\right) \\ & = \sqrt{n}\left(\sum_{j=1}^\infty \frac{\Gamma(3/2)}{j! \Gamma(3/2 - j)} n^{-j}\right) \\ & \approx \frac{1}{2\sqrt{n}} - \mathcal{O}(n^{-3/2}) \\ & \rightarrow 0. \end{align}$$

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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.

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