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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3$\begingroup$ Hint: Rationalize and it suffices to show that the $\sqrt{n+1}+\sqrt{n}\to\infty$ $\endgroup$– MoyaDec 20, 2015 at 1:45
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1$\begingroup$ Closely related, but not a duplicate (because the question only allows one solving method): Show that $\lim_{n\to \infty} (\sqrt{n+1} - \sqrt n) = 0$ using the definition of a limit $\endgroup$– SilverfishDec 20, 2015 at 13:19
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$\begingroup$ See also: Show that $\sqrt{n+1}-\sqrt{n}\to0$ $\endgroup$– Martin SleziakApr 18, 2017 at 3:36
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4 Answers
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$$ \sqrt{n+1}-\sqrt{n} = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}} $$
answered Dec 20, 2015 at 1:45
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$\begingroup$ Shouldn't the last part be $\leq \frac{1}{2\sqrt{n+1}}$? $\endgroup$ Dec 20, 2015 at 1:57
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4$\begingroup$ @cheesyfluff That is not true since $2\sqrt{n+1} > \sqrt{n+1} + \sqrt{n}$.. $\endgroup$ Dec 20, 2015 at 2:00
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2$\begingroup$ Can you explain how you realized you're supposed to use conjugates? It's not really enlightening otherwise. $\endgroup$ Dec 20, 2015 at 5:19
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2$\begingroup$ For squeezing, we could use from your results $$ \frac{1}{2\sqrt{n+1}}<\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}}$$ $\endgroup$ Dec 20, 2015 at 5:30
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1$\begingroup$ @Mehrdad I try conjugates whenever I see differences of the form $\sqrt{a}-\sqrt{b}$ and that $a-b$ is simple. $\endgroup$ Dec 20, 2015 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 $0 < \displaystyle \sqrt{x+1} - \sqrt{x} < \frac1{2\sqrt x}$.
answered Dec 20, 2015 at 2:18
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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.
answered Dec 20, 2015 at 1:58