Convergence of the series $\sum \frac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$ 
To prove that nature of the following series : $$\sum \dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$$
  they use in solution manual :


My questions:


*

*I don't know how to achieve ( * ) could someone complete my attempts for ( * )  and is it correct if i use (**) to prove that the series is convergent : 


$$\fbox{$\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}-\dfrac{(-1)^{n}}{n}+O\left(\dfrac{1}{n^{\frac{4}{3}}}\right)$}\quad (*)$$
My thoughts :
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}
\end{align}

note that :
  $$(1+x)^{\alpha}=1+\alpha x+O(x^{2})$$

\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1-\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)+O\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)^{2} \right)\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} -\dfrac{(-1)^{n}}{n}+\dfrac{(-1)^{n}}{n^{\frac{4}{3}}} +\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}\times O\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)^{2} \\
&=\ldots\ldots \\
&= \mbox{ I'm stuc here i  hope someone complete my attempts } 
\end{align} 
Or i should use :

note that :
  $$(1+x)^{\alpha}=1+O(x)$$

\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+O\left( \dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}\right) \right)\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{1}{n}+\dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right) \\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right) \\
\end{align}
$$\fbox{$\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} =\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right)$}\quad (**) $$
 A: Fix $N>0
 $ and $N
 $ even (you can also take $N
 $ odd, the proof is essentially the same). We have $$\sum_{n=1}^{N}\frac{\left(-1\right)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+\left(-1\right)^{n}}=\sum_{n=1}^{N/2}\frac{1}{\left(2n\right)^{\frac{2}{3}}+\left(2n\right)^{\frac{1}{3}}+1}-\sum_{n=1}^{N/2-1}\frac{1}{\left(2n-1\right)^{\frac{2}{3}}+\left(2n-1\right)^{\frac{1}{3}}-1}
 $$ $$=\sum_{n=1}^{N/2}\frac{\left(2n-1\right)^{2/3}-\left(2n\right)^{2/3}+\sqrt[3]{2n-1}-\sqrt[3]{2n}-2}{\left(\left(2n\right)^{\frac{2}{3}}+\left(2n\right)^{\frac{1}{3}}+1\right)\left(\left(2n-1\right)^{\frac{2}{3}}+\left(2n-1\right)^{\frac{1}{3}}-1\right)}-\frac{1}{\left(N-3\right)^{\frac{2}{3}}+\left(N-3\right)^{\frac{1}{3}}-1}
 $$ $$=\sum_{n=1}^{N}\frac{\left(-1\right)^{n}}{n^{2/3}\left(1+O\left(1\right)\right)}+\sum_{n=1}^{N}\frac{\left(-1\right)^{n}}{n\left(1+O\left(1\right)\right)}$$ $$-\sum_{n=1}^{N/2}\frac{2}{n^{4/3}\left(1+O\left(1\right)\right)}-\frac{1}{\left(N-3\right)^{\frac{2}{3}}+\left(N-3\right)^{\frac{1}{3}}-1}
 $$ and so the series converges.
A: $\dfrac{1}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}-\dfrac{1}{n^{\frac{2}{3}}}+\dfrac{1}{n} = \dfrac{-n(-1)^n+n + n^{2/3}(-1)^n}{n^{5/3}(n^{2/3}+n^{1/3}+(-1)^n}$
The right hand side has $n$ in the numerator and $n^{7/3}$ in the denominator and hence is $O\left(\frac{1}{n^{4/3}}\right)$
A: Let $f(n)=\frac{1}{n^{2/3}+n^{1/3}+A}$ and consider $\phi_m(n)=n^{-m/3}$ for $m=2,3,...$ Then you should be after an asymptotic expnasion for $f$ of the form $$f(n)=\sum a_m\phi_m(n).$$ We have $$a_2=\lim_{n\to\infty}\frac{f(n)}{\phi_2(n)}=1$$ and $$a_3=\lim_{n\to\infty}\frac{f(n)-a_2\phi_2(n)}{\phi_3(n)}=-1$$ and $$a_3=\lim_{n\to\infty}\frac{f(n)-a_2\phi_2(n)-a_3\phi_3(n)}{\phi_4(n)}=1-A$$ 
etc...
and as such we obtain $$f(n)=\frac{1}{n^{2/3}}-\frac1n+(1-A)\frac{1}{n^{4/3}}+O(n^{5/3})$$
