Show that $\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}n+\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$ 
How can i prove that $$\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}{n} +\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\tag{$*$}$$

using the following method :

note that : 
  $(1+x)^{\alpha}=1+\alpha x+\mathcal{O}(x^{2})\quad ( x\to 0) $

\begin{align}
\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}&=\frac{(-1)^n}{n} \left(1+\frac{(-1)^n\sqrt{n+1}}{n} \right)^{-1}\\
&\text{since } \sqrt{n+1}\sim \sqrt{n} \text{ then } \lim_{n\to+\infty}\frac{\sqrt{n+1}}{n}=0 \\
&\sim \frac{(-1)^n}{n}\left(1+\frac{(-1)^n\sqrt{n}}{n} \right)^{-1}\\
&= \frac{(-1)^{n}}n\left(1+\mathcal{O}\left( \frac{(-1)^n\sqrt{n}}{n}\right) \right)\\
&= \frac{(-1)^n}{n}+\mathcal{O}\left( \dfrac{(-1)^{2n}\sqrt{n}}{n^2}\right) \\
&=\frac{(-1)^n}{n}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)
\end{align}


*

*AM i right ?  

 A: One may write, as $n \to \infty$,
$$
\begin{align}
\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}&=\frac{(-1)^n}{n} \left(1+\frac{(-1)^n\sqrt{n+1}}{n} \right)^{-1}\\
&= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}\cdot\sqrt{1+\frac1n}\: \right)^{-1}\\
&= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}\left(1+\frac1{2n}+\mathcal{O}\left(\dfrac{1}{n^2}\right)\right)\: \right)^{-1}\\
&=  \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\: \right)^{-1}\\
&=  \frac{(-1)^n}{n}\left(1-\frac{(-1)^n}{\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\: \right)\\
&=\frac{(-1)^n}{n}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)
\end{align}
$$ as wanted.
A: I would initially ignore the
$(-1)^n$
and do
$\begin{array}\\
\frac{1}{n+(-1)^n\sqrt{n+1}}-\frac{1}{n}
&=\frac{n-(n+(-1)^n\sqrt{n+1})}{n(n+(-1)^n\sqrt{n+1})}\\
&=\frac{-(-1)^n\sqrt{n+1}}{n(n+(-1)^n\sqrt{n+1})}\\
&=\frac1{n^{3/2}}\frac{(-1)^{n+1}\sqrt{1+1/n}}{1+(-1)^n\sqrt{1/n+1/n^2})}\\
\text{so}\\
\big|\frac{1}{n+(-1)^n\sqrt{n+1}}-\frac{1}{n}\big|
&=\frac1{n^{3/2}}\frac{\sqrt{1+1/n}}{1+(-1)^n\sqrt{1/n+1/n^2})}\\
&=O(\frac1{n^{3/2}})\\
\end{array}
$
since
$\sqrt{1+1/n} \to 1$
and
$\sqrt{1/n+1/n^2} \to 0$.
