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So, the series is: $$\sum_{n=1}^\infty\frac{\sqrt{n+2}-\sqrt{n}}{n^k}$$ What I did is: $$a_n=\frac{\sqrt{n+2}-\sqrt{n}}{n^k}*\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}=\frac{2}{n^k(\sqrt{n+2}+\sqrt{n})}$$ I'm not sure what to do next, which criteria should I use?

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  • $\begingroup$ What is $\;k\;$ ? $\endgroup$ – Timbuc Apr 6 '15 at 17:55
  • $\begingroup$ Are you supposed to find values of $k$ for which the series will converge? $\endgroup$ – graydad Apr 6 '15 at 17:55
  • $\begingroup$ so $a_n\sim n^{-(k+\frac 12)}$ hence $a_n$ is convegent if and only if $\cdots$ $\endgroup$ – Elaqqad Apr 6 '15 at 17:56
  • $\begingroup$ you can do limit comparison with $\frac 1{n^{k+1/2}}$ $\endgroup$ – abel Apr 6 '15 at 17:57
  • $\begingroup$ k is probably a constant for which I have to test the convergence in different cases. $\endgroup$ – A6SE Apr 6 '15 at 18:08
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Note

$$\frac{1}{2n^{k+1/2}} < a_n < \frac{2}{n^{k+1/2}}$$

for all $n \ge 1$. Thus $\sum\limits_{n = 1}^\infty a_n$ converges if and only if $\sum\limits_{n = 1}^\infty n^{-(k+1/2)}$ converges. Now use the $p$-test.

Remark: To get the lower bound $\frac{1}{2n^{k + 1/2}} < a_n$, consider that $$\sqrt{n+2} + \sqrt{n} <2\sqrt{n+2} < 2\sqrt{n + 3n} = 2\sqrt{4n} = 4\sqrt{n}.$$

This implies $a_n > \frac{2}{n^k(4\sqrt{n})} = \frac{1}{2n^{k+1/2}}$.

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  • $\begingroup$ Your first expression, shouldn't it be > instead of <? $\endgroup$ – A6SE Apr 6 '15 at 19:19
  • $\begingroup$ No. I gave an explanation in the remark. $\endgroup$ – kobe Apr 6 '15 at 19:21
  • $\begingroup$ I understand most of it, but still find some things unclear. How did you get the upper bound and do you need both bounds to test the convergence? $\endgroup$ – A6SE Apr 6 '15 at 19:32
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    $\begingroup$ We need both bounds. If $\sum\limits_{n = 1}^\infty n^{-(k+1/2)}$ converges, then since $0 < a_n < 2n^{-(k+1/2)}$ for all $n$, direct comparison shows $\sum\limits_{n = 1}^\infty a_n$ converges. On the other hand, if $\sum\limits_{n = 1}^\infty n^{-(k+1/2)}$ diverges, then since $a_n > 0.5n^{-(k+1/2)}$ for all $n$, direct comparison shows $\sum\limits_{n = 1}^\infty a_n$ diverges. Hence $\sum\limits_{n = 1}^\infty a_n$ converges iff $\sum\limits_{n = 1}^\infty n^{-(k+1/2)}$ converges. $\endgroup$ – kobe Apr 6 '15 at 19:45
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    $\begingroup$ All it means is you don't have enough information to determine whether of not $\sum\limits_{n = 1}^\infty a_n$ diverges. $\endgroup$ – kobe Apr 6 '15 at 19:59
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$\sqrt{n+2}+\sqrt{n}\sim 2\sqrt{n}$ and use the comparison test.

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Hint:

$$\frac1{n^k(\sqrt{n+2}+\sqrt n)}\le\frac1{n^k\cdot2\sqrt n}=\frac12\frac1{n^{k+\frac12}}$$

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