I would like to show that : $$\dfrac{(-1)^{n}}{\left(\ln(n)+(-1)^{n}\right)^{2}}=\dfrac{(-1)^{n}}{\ln^{2}(n)}+v_n\quad \left( v_n\sim -\dfrac{2}{\ln^{3}(n)} \right)\\ $$ by starting from the left side and get the right side
My proof:
\begin{align*} \dfrac{(-1)^{n}}{\left(\ln(n)+(-1)^{n}\right)^{2}}&= \dfrac{(-1)^{n}}{\ln^{2}(n)}\left(1+\dfrac{(-1)^n}{\ln(n)} \right)^{-2} \\ &=\dfrac{(-1)^{n}}{\ln^{2}(n)}\left(1-2\dfrac{(-1)^n}{\ln(n)} +\mathcal{O}\left( \dfrac{1}{\ln(n)} \right) \right)\\ &=\dfrac{(-1)^{n}}{\ln^{2}(n)}-\dfrac{2}{\ln^{3}(n)} +\mathcal{O}\left( \dfrac{1}{\ln^{3}(n)} \right) \\ &=\dfrac{(-1)^{n}}{\ln^{2}(n)}+v_n\quad \left( v_n\sim -\dfrac{2}{\ln^{3}(n)} \right)\\ \end{align*}
$$-\dfrac{2}{\ln^{3}(n)} +\mathcal{O}\left( \dfrac{1}{\ln^{3}(n)} \right) \sim -\dfrac{2}{\ln^{3}(n)}$$ since :
- $$\left| \dfrac{-\dfrac{2}{\ln^{3}(n)} +\mathcal{O}\left( \dfrac{1}{\ln^{3}(n)} \right) }{-\dfrac{2}{\ln^{3}(n)}}\right|=\left|1+\dfrac{\mathcal{O}\left( \dfrac{1}{\ln^{3}(n)} \right)}{-\dfrac{2}{\ln^{3}(n)}} \right|\leq 1+\dfrac{1}{2}$$ i can't prove that goes to 1 when n tend to infinity
Update:
Could someone please prove $$-\dfrac{2}{\ln^{3}(n)} +\mathcal{O}\left( \dfrac{1}{\ln^{3}(n)} \right) \sim -\dfrac{2}{\ln^{3}(n)}$$ Is it true that : $$-\dfrac{2}{\ln^{3}(n)} +\mathcal{o}\left( \dfrac{1}{\ln^{3}(n)} \right) \sim -\dfrac{2}{\ln^{3}(n)} \implies -\dfrac{2}{\ln^{3}(n)} +\mathcal{O}\left( \dfrac{1}{\ln^{3}(n)} \right) \sim -\dfrac{2}{\ln^{3}(n)}$$
Replace $\mathcal O\Bigl(\dfrac1{\ln n}\Bigr)$, which is not false, but doesn't bring any information here, with the correct $ \;o\Bigl(\dfrac1{\ln n}\Bigr)$. You'll get $$-\frac{2}{\ln^{3}n} +\mathcal{o}\Bigl( \frac{1}{\ln^{3}n} \Bigr) \sim -\frac{2}{\ln^{3}n}$$ by definition.