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My attempt:

$\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n^x} = \sum_{n=1}^\infty\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}$

From here, I've tried using the ratio test, but haven't been able to simplify it in a clever way. I can bound the above sum by $\sum_{n=1}^\infty\frac{1}{n^x}$, but this won't show all $x\in\mathbb{R}$ that cause the sum to converge. Basically, I am stuck now.

Any help appreciated!

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    $\begingroup$ For which $x$ does $\sum \frac{1}{n^x\sqrt{n}}$ converge? $\endgroup$ – Daniel Fischer Feb 6 '17 at 20:32
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HINT:

$$\frac{1}{n^x\left(\sqrt{n+1}+\sqrt{n}\right)}< \frac{1}{n^x(2\sqrt{n})}=\frac12 \frac{1}{n^{x+1/2}}$$

and

$$\frac{1}{n^x\left(\sqrt{n+1}+\sqrt{n}\right)}> \frac{1}{n^x(3\sqrt{n})}=\frac13 \frac{1}{n^{x+1/2}}$$

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  • $\begingroup$ So for these examples, do we have that $\frac{1}{2}\frac{1}{n^{x+1/2}}$ converges for $x>1/2$ and $\frac{1}{3}\frac{1}{n^{x+1/2}}$ diverges for $x<1/2$, so the sum converges for all $x>1/2$? $\endgroup$ – Jess Feb 7 '17 at 12:58
  • $\begingroup$ @jess Yes. The series converges when $x>1/2$ and diverges elsewhere. $\endgroup$ – Mark Viola Feb 7 '17 at 15:15
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As this is a series with positive terms, I would use equivalents:

$\sqrt{n+1}-\sqrt{n}\sim_\infty\dfrac1{2\sqrt n}$, so $$\frac{\sqrt{n+1}-\sqrt{n}}{n^x}\sim_\infty\dfrac1{2\, n^{\,x+\frac12}},$$ which converges if and only if $x+\frac12>1$, i.e. $x>\frac12$, and diverges if and only if $x\le \frac12$.

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    $\begingroup$ You probably meant $n^{x+1/2}$ instead of $\sqrt{n}^{x+1/2}$ $\endgroup$ – kingW3 Feb 6 '17 at 21:02
  • $\begingroup$ Oh! yes. Thanks for pointing it. Fixed now. $\endgroup$ – Bernard Feb 6 '17 at 21:15

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