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How to prove convergence of the following?

$$\sum_{k=1}^n {\sqrt{k^3+1}-\sqrt {k^3-1}}$$

Thanks!

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    $\begingroup$ $k$ begin from $1$. $\endgroup$
    – Nosrati
    Sep 15, 2017 at 2:12
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    $\begingroup$ yes sorry ill change it $\endgroup$ Sep 15, 2017 at 2:12

2 Answers 2

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hint:$$\sum_{k=1}^n (\sqrt{k^3+1}-\sqrt {k^3-1})=\\\sum_{k=1}^n (\sqrt{k^3+1}-\sqrt {k^3-1})\frac{\sqrt{k^3+1}+\sqrt {k^3-1}}{\sqrt{k^3+1}+\sqrt {k^3-1}} =\\\sum_{k=1}^n \frac{2}{\sqrt{k^3+1}+\sqrt {k^3-1}}\sim \sum_{k=1}^n \frac{2}{2\sqrt{k^3}}$$

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$\sqrt{k^3+1}-\sqrt {k^3-1}=\frac{(\sqrt{k^3+1}-\sqrt {k^3-1})(\sqrt{k^3+1}+\sqrt {k^3-1})}{(\sqrt{k^3+1}+\sqrt {k^3-1})}=\frac{2}{\sqrt{k^3+1}+\sqrt {k^3-1}}\leq \frac{2}{\sqrt{k^3+1}}\leq \frac{2}{\sqrt{k^3}}=\frac{2}{k^{1+\frac{1}{2}}}$

Then apply comparison test.

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