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How would I go about proving that $\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}=0$? I have tried to use Squeeze theorem but have not been able to come up with bounds that converge to zero. Additionally, I don't think that converting to polar is possible here.

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$$ \sqrt{n+1}-\sqrt{n} = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}} $$

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  • $\begingroup$ Shouldn't the last part be $\leq \frac{1}{2\sqrt{n+1}}$? $\endgroup$ – cheesyfluff Dec 20 '15 at 1:57
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    $\begingroup$ @cheesyfluff That is not true since $2\sqrt{n+1} > \sqrt{n+1} + \sqrt{n}$.. $\endgroup$ – Henricus V. Dec 20 '15 at 2:00
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    $\begingroup$ Can you explain how you realized you're supposed to use conjugates? It's not really enlightening otherwise. $\endgroup$ – user541686 Dec 20 '15 at 5:19
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    $\begingroup$ For squeezing, we could use from your results $$ \frac{1}{2\sqrt{n+1}}<\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}}$$ $\endgroup$ – Claude Leibovici Dec 20 '15 at 5:30
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    $\begingroup$ @Mehrdad I try conjugates whenever I see differences of the form $\sqrt{a}-\sqrt{b}$ and that $a-b$ is simple. $\endgroup$ – Henricus V. Dec 20 '15 at 5:52
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Using conjugate multiplication can be quite useful in cases like that: $$\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n}){\sqrt{n+1}+\sqrt{n}\over \sqrt{n+1}+\sqrt{n}}={(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})\over \sqrt{n+1}+\sqrt{n}}={(n+1)-n\over \sqrt{n+1}+\sqrt{n}}={1\over \sqrt{n+1}+\sqrt{n}}$$. The last result is pretty easy to work with.

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One way is by using the mean value theorem. Specifically, let $f(x) = \sqrt x$. Then, for each $x > 0$, we know that $\displaystyle f(x+1) - f(x) = \frac{f(x+1) - f(x)}{(x+1) - x} = f'(c)$ for some $c$ in the interval $(x, x+1)$. Since $\displaystyle f'(x) = \frac1{2\sqrt x}$ is strictly decreasing we conclude that $0 < \displaystyle \sqrt{x+1} - \sqrt{x} < \frac1{2\sqrt x}$.

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Note that for positive $n$ we have $$\sqrt{n}\lt \sqrt{n+1}\lt \sqrt{n}+\frac{1}{2\sqrt{n}}.\tag{1}$$ The second inequality in (1) holds because $$\left(\sqrt{n}+\frac{1}{2\sqrt{n}}\right)^2=n+1+\frac{1}{4n}\gt n+1.$$ It follows from (1) that $$0\lt \sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}.$$ Now Squeeze.

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