Why we have that $$\begin{align*} \frac{(-1)^n}{\sqrt{n}}\left(1+\frac{(-1)^n}{\sqrt{n}} \right)^{-1} & =\frac{(-1)^n}{\sqrt{n}}\left(1-\frac{(-1)^n}{\sqrt{n}}+O\left(\frac{1}n\right) \right) \tag{$E_1$} \\ \end{align*}$$ insted of $$\left(1-\frac{(-1)^n}{\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right) \right) \tag{$E_2$} ?$$ Is that because we have $\frac{1}{\sqrt{n}}=o\left(\frac{1}{n}\right)$? But this is wrong!
On
By the Taylor series expansion, as $u \to 0$, we have $$ \frac1{1+u}=1-u+O(u^2) $$ giving, as $n \to \infty$, $$ \frac1{1+\dfrac{(-1)^n}{\sqrt{n}}}=1-\dfrac{(-1)^n}{\sqrt{n}}+O\left(\dfrac{(-1)^n}{\sqrt{n}}\right)^2=1-\dfrac{(-1)^n}{\sqrt{n}}+O\left(\dfrac1n\right), $$ and $$ \dfrac{(-1)^n}{\sqrt{n}}\frac1{1+\dfrac{(-1)^n}{\sqrt{n}}}=\dfrac{(-1)^n}{\sqrt{n}}-\dfrac1n+O\left(\dfrac1{n^{3/2}}\right). $$
By Taylor expansion, \begin{align*} \frac{1}{1+\frac{(-1)^n}{\sqrt{n}}} &= 1+ \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n} + o\left(\frac{1}{n}\right)\\ &=1+ \frac{(-1)^n}{\sqrt{n}} + O\left(\frac{1}{n}\right). \end{align*}