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I already tried the ratio and root criterion, but it didn't get me anywhere. I'd be thrilled if you had any suggestions.

(also applied the third binomial formula, so it would "look nicer", maybe it helps $\sum_{n=1}^{\infty}{\frac{\sqrt{n+1}-\sqrt{n}}{n^q}}=\sum_{n=1}^{\infty}{\frac{1}{n^q(\sqrt{n+1}+\sqrt{n})}}$)

thanks and greetings!

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Clearly $$ \frac{1}{2(n+1)^{q+{1/2}}}<\frac{\sqrt{n+1}-\sqrt{n}}{n^q}=\frac{1}{(\sqrt{n+1}+\sqrt{n})\cdot n^q}<\frac{1}{2n^{q+{1/2}}}, $$ and $\sum_{n=1}^\infty\frac{1}{n^{q+1/2}}$, equivalently $\sum_{n=1}^\infty\frac{1}{(n+1)^{q+1/2}}$, converge, if and only if $q+1/2>1$, equivalently $q>1/2$.

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  • $\begingroup$ Equivalently $q>-1/2$? $\endgroup$ – user Nov 29 '18 at 15:47
  • $\begingroup$ Thanks! I corrected it! $\endgroup$ – Yiorgos S. Smyrlis Nov 29 '18 at 15:49
  • $\begingroup$ You are welcome! minor typo of course! $\endgroup$ – user Nov 29 '18 at 15:50
  • $\begingroup$ thx a lot, that helps, I assume you conclude the convergence by the direct comparison test. Another tiny "problem": You showed that for those q that suffice $q>-\frac{1}{2}$ the series must converge. But how do we know there is no q, for which $\frac{1}{2n^{q+0.5}}$ diverges but $\frac{1}{\sqrt{n+1}+\sqrt{n}}$ still converges. Do I have to show that or is it obvious/trivial? $\endgroup$ – DDevelops Nov 29 '18 at 17:05
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Hint 1:

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

Hint 2 Limit compare ${\frac{1}{n^q(\sqrt{n+1}+\sqrt{n})}}$ to ${\frac{1}{n^q\sqrt{n}}}$

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From here

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

we have that

$$\frac{1}{n^q(\sqrt{n+1}+\sqrt{n})}\sim \frac{1}{2n^{q+\frac12}}$$

then refer to limit comparison test to conlcude that $q+\frac12>1$.

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