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Find $p$ that makes $\sum\limits_{n=1}^{\infty} {\dfrac{(-1)^{n-1}}{(\sqrt{n}+(-1)^{n-1})^p}}$ converges. Which $p$ makes the series converges absolutely?

I think that it converges for $p>0$, can I use: ${\dfrac{1}{{{{\left( {\sqrt{n} + {{\left( { - 1} \right)}^{n - 1}}} \right)}^p}}}} \sim \dfrac{1}{{{n^{\frac{p}{2}}}}}$ to conclude the series converges absolutely for $p>2$?

Thanks in advance!

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    $\begingroup$ The series diverges for $p= 1$ , since $$\sum_{n=2}^N \frac{(-1)^{n-1}}{\sqrt{n} + (-1)^{n-1}} = \sum_{n=2}^N \frac{(-1)^{n-1}\sqrt{n}}{n -1} - \sum_{n=2}^N \frac{1}{n -1},$$ and the first series on the RHS converges (Dirichlet) but the second series diverges. $\endgroup$ – RRL Mar 23 '17 at 17:34
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    $\begingroup$ Converges for $p>1$, diverges for $p\leq 1$. $\endgroup$ – i707107 Mar 23 '17 at 17:39
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    $\begingroup$ ...and converges absolutely for $p>2$ as pointed by OP. $\endgroup$ – i707107 Mar 23 '17 at 17:53
  • $\begingroup$ Thanks, I get it. Just use Taylor series to expand a factor and we will get the result. $\endgroup$ – dienhosp3 Mar 24 '17 at 7:48

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