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wan't know how to start this problem, tried to multiply by $\sqrt{n}$, but it does not work. Can someone help me with that? Thanks. Here is the inequality. $$2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$$

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

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  • $\begingroup$ Thank you for the hint! $\endgroup$ – aeon Jun 22 '17 at 3:21
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By MVT,

$$\sqrt {n}-\sqrt {n-1}=\frac {1}{2\sqrt {c}} $$

$$\sqrt {n+1}-\sqrt {n}=\frac {1}{2\sqrt {d}} $$ with $$n-1 <c <n<d <n+1 . $$

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