I would like to show that : $$\dfrac{(-1)^{n}}{\cos(n)+n^{\tfrac{3}{4}}}=\dfrac{(-1)^{n}}{n^{\tfrac{3}{4}}}+\mathcal{O}\left( \dfrac{1}{n^{\tfrac{3}{2}}}\right) $$ by starting from the left side and get the right side
My Proof:
Note that : $$\left(1+x\right)^{\alpha}=1+\mathcal{O}\left( x\right) \quad x \to 0$$ $$\left|\left(\dfrac{\cos(n)}{n^{\tfrac{3}{4}}} \right)\right|\leq \left(\dfrac{1}{n^{\tfrac{3}{4}}} \right) \underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow }0 $$
\begin{align*} \dfrac{(-1)^{n}}{\cos(n)+n^{\tfrac{3}{4}}}&=\dfrac{(-1)^{n}}{n^{\tfrac{3}{4}}}\left(1+\dfrac{\cos(n)}{n^{\tfrac{3}{4}}} \right)^{-1} \\ &= \dfrac{(-1)^{n}}{n^{\tfrac{3}{4}}}\left(1+\mathcal{O}\left(\dfrac{\cos(n)}{n^{\tfrac{3}{4}}} \right) \right)\\ &=\dfrac{(-1)^{n}}{n^{\tfrac{3}{4}}}+\mathcal{O}\left( \dfrac{\cos(n)}{n^{\tfrac{3}{2}}}\right) \\ &=\dfrac{(-1)^{n}}{n^{\tfrac{3}{4}}}+\mathcal{O}\left( \dfrac{1}{n^{\tfrac{3}{2}}}\right) \end{align*}
since $$\left|\dfrac{\left( \dfrac{\cos(n)}{n^{\tfrac{3}{2}}}\right) }{\left( \dfrac{1}{n^{\tfrac{3}{2}}}\right) }\right| \leq 1 \implies \mathcal{O}\left( \dfrac{\cos(n)}{n^{\tfrac{3}{2}}}\right) =\mathcal{O}\left( \dfrac{1}{n^{\tfrac{3}{2}}}\right) $$
- Is my proof correct