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

I think that it converges for $p&gt;0$, 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 $p&gt;2$?

Thanks in advance!
Answered in the comments, but let's do it explicitly.

$\sum\limits_{n=1}^\infty a_n$ converges if and only if $\lim\limits_{n\to\infty}a_n=0$ and $\sum\limits_{n=1}^\infty(a_{2n-1}+a_{2n})$ converges.

For $a_n=(-1)^{n-1}\big(\sqrt{n}+(-1)^{n-1}\big)^{-p}$, this translates into [$p&gt;0$ and] $\color{blue}{p&gt;1}$, since $$a_{2n-1}+a_{2n}=f(1/\sqrt{2n}),\quad f(x):=x^p\big((\sqrt{1-x^2}+x)^{-p}-(1-x)^{-p}\big),$$ so that $\lim\limits_{x\to0^+}x^{-p-1}f(x)=-2p$, and we're done by the comparison test, with $\sum\limits_{n=1}^\infty n^{-(p+1)/2}$.
For $n\geq 1$, let $u_n=\frac{(-1)^{n-1}}{(\sqrt{n}+(-1)^{n-1})^p}$

1) For $p&lt;0$, we have

$u_n=(\sqrt{n})^{-p}(-1)^{n-1}\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^{-p}$

So $u_n \sim (-1)^{n-1}(\sqrt{n})^{-p}$ ($n\rightarrow +\infty$)

so $\lim_{n\rightarrow +\infty} |u_n|=+\infty$

Thus $(u_n)$ doesn't tend to 0.

The series is divergent.

2) For $p=0$, $u_n=(-1)^{n-1}$

We know that the series is divegent.

3) Suppose $p&gt;0$

We have $u_n=\frac{(-1)^{n-1}}{(\sqrt{n})^p}.\frac{1}{\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^p}$

We deduce that 

$|u_n| \sim \frac{1}{n^{p/2}}$ ($n\rightarrow +\infty$)

So the series $\sum u_n$ is absolutely convergent if and only if $p/2&gt;1$, if and only if $p&gt;2$.

But absolutely convergent implies convergent

4) Suppose $0&lt;p\leq 2$

We make a series expansion for $n\geq 2$

$\frac{1}{\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^p}=\sum_{k=0}^{+\infty} {-p \choose k}\left(\frac{(-1)^{n-1}}{\sqrt{n}}\right)^k$

Multiplying by $\frac{(-1)^{n-1}}{n^{p/2}}$, the power over $n$ is $\frac{p}{2}
+\frac{k}{2}=\frac{k+p}{2}$

and $\frac{k+p}{2}&gt;1\Leftrightarrow k&gt;2-p$

So for $k\geq 2$ there will be absolute convergence

$\frac{1}{\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^p}=1-p\frac{(-1)^{n-1}}{\sqrt{n}}+O\left(\frac{1}{n}\right)$

$\frac{(-1)^{n-1}}{(\sqrt{n})^p}.\frac{1}{\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^p}=\frac{(-1)^{n-1}}{n^{p/2}}-p\frac{1}{n^{(p+1)/2}}+O\left(\frac{1}{n^{p/2+1}}\right)$

The last term is of the form $\frac{b_n}{n^{p/2+1}}$ with $(b_n)$ a sequence bounded by a real number $M$. However, the series $\sum \frac{M}{n^{p/2+1}}$ is absolutely convergent because $p/2+1&gt;1$, so $\sum \frac{b_n}{n^{p/2+1}}$ is absolutely convergent.

The series $\frac{(-1)^{n-1}}{n^{p/2}}$ is an alternate Riemman series with $p/2&gt;0$ so is convergent.

Finally the Riemman series $\sum \frac{1}{n^{(p+1)/2}}$ with positive terms is convergent if and only if $(p+1)/2&gt;1$ i.e. $p&gt;1$.

We can conclude that the series $\sum u_n$ is convergent if $1&lt;p\leq 2$ as sum of convergent series.

and divergent if $0&lt;p\leq 1$ as the sum of a divergent series and others convergent.