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$$\sum(2\sqrt{n+1}-\sqrt{n+2}-\sqrt{n})$$

I have tried comparison test but gives the best I get was $\dfrac{1}{\sqrt{n}}$.

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    $\begingroup$ You should compute some partial sums and see what you get. The sum telescopes so you should get something fairly simple (the exact expression depends on the lower limit of the sum—I'm assuming $n = 1$, but it could be something else). $\endgroup$ – Brian Tung Jan 31 '17 at 18:03
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Hint: Write as $$(\sqrt{n+1}-\sqrt{n})+(\sqrt{n+1}-\sqrt{n+2})$$ and combine each term using the conjugate.

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  • $\begingroup$ gives me only $\frac{1}{n}$ , so still nothing $\endgroup$ – UfmdFkiF Jan 31 '17 at 18:32
  • $\begingroup$ @TheMeff Not true, if you proceed carefully as vadim123 indicated, you should get that the general term is $$\frac2{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+2}+\sqrt{n+1})(\sqrt{n+1}+\sqrt{n})}$$ from which the conclusion is direct. $\endgroup$ – Did Mar 5 '17 at 12:41
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$$2\sqrt{n+1}-\sqrt{n+2}-\sqrt n=(\sqrt{n+1}-\sqrt n)-(\sqrt{n+2}-\sqrt{n+1})=f(n)-f(n+1)$$ where $f(m)=\sqrt{m+1}-\sqrt m$

Partial Sum $$\sum_{n=1}^r(2\sqrt{n+1}-\sqrt{n+2}-\sqrt n)=\sum_{n=1}^r(f(n)-f(n+1))=\cdots=f(1)-f(r+1)$$

Now $f(m)=\sqrt{m+1}-\sqrt m=\dfrac{m+1-m}{\sqrt m+\sqrt{(m+1)}}$

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Just follow Brian Tung's hint: $$\sum_{n=1}^{N}\left[(\sqrt{n+1}-\sqrt{n})-(\sqrt{n+2}-\sqrt{n+1})\right] = (\sqrt{2}-\sqrt{1})-(\sqrt{N+2}-\sqrt{N+1}) $$ then notice that $0\leq\sqrt{N+2}-\sqrt{N+1}=\frac{1}{\sqrt{N+1}+\sqrt{N+2}}\leq\frac{1}{2\sqrt{N}}$.

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