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So, I am trying to show if $\sum_{n=1}^\infty \frac{1}{\sqrt{n+2}+\sqrt{n}}$ diverges or converges through the comparison test. However, I am having difficulty finding something to compare it to. I know that it diverges, but I am unsure of a smaller series that also diverges. Thanks in advance.

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Hint: $$\frac{1}{\sqrt{n+2}+\sqrt{n}}>\frac{1}{2\sqrt{n+2}}.$$

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  • $\begingroup$ Yeah, this was the route I went, but how is $\frac{1}{2\sqrt{n+2}}$ divergent? $\endgroup$ – RXY15 Nov 3 '14 at 4:34
  • $\begingroup$ @RXY15: $$\frac1{2\sqrt{n+2}}\ge\frac1{2\sqrt{2n}}=\frac1{2\sqrt2}\frac1{\sqrt n}\;$$ $\endgroup$ – Timbuc Nov 3 '14 at 4:35
  • $\begingroup$ Ok, and then from there $\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{n}} \geq \frac{1}{2\sqrt{2}}\frac{1}{n}$, which diverges. $\endgroup$ – RXY15 Nov 3 '14 at 4:37
  • $\begingroup$ @RXY15: that is one such way (or if you have the integral test). $\endgroup$ – Clayton Nov 3 '14 at 4:39
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Alternatively, we have: $\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n+2} + \sqrt{n}} = \dfrac{1}{2}\displaystyle \sum_{n=1}^\infty \left(\sqrt{n+2}-\sqrt{n}\right) \geq \dfrac{1}{2}\cdot \displaystyle \lim_{n \to \infty} \sum_{k=1}^n \left(\sqrt{k+1}-\sqrt{k}\right) = \dfrac{1}{2}\cdot \displaystyle \lim_{n \to \infty} \left(\sqrt{n+1} - 1\right) = \infty$

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  • $\begingroup$ In the term after the inequality, I believe you should have $\sqrt{k+2}-\sqrt{k}$ as the summand. $\endgroup$ – Clayton Nov 3 '14 at 13:36

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