Show $ \frac{(-1)^{n}}{n-\ln(n)}=\frac{(-1)^{n}}{n}+\mathcal{O}\left(\frac{\ln(n)}{n^{2}} \right) $

Question

I would like to show that : $$ \dfrac{(-1)^{n}}{n-\ln(n)}=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right) $$ by starting from the left side and get the right side

My proof:

note that :

• $$\left( 1+x \right)^{\alpha}=1+\alpha x+\mathcal{O}\left( x^2\right) $$
• $$ \dfrac{\ln(n)}{n}\underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow}0$$

\begin{align*} \dfrac{(-1)^{n}}{n-\ln(n)}&=\dfrac{(-1)^{n}}{n}\left(1-\dfrac{\ln(n)}{n} \right)^{-1} \\ &=\dfrac{(-1)^{n}}{n}\left(1+\dfrac{\ln(n)}{n}+\mathcal{O}\left(\dfrac{\ln^2(n)}{n^2} \right) \right) \\ &=\dfrac{(-1)^{n}}{n}+\dfrac{(-1)^{n}\ln(n)}{n^2}+\mathcal{O}\left(\dfrac{\ln^2(n)}{n^3} \right) \\ &=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right) \\ \end{align*}

Since : $$\dfrac{(-1)^{n}\ln(n)}{n^2}+\mathcal{O}\left(\dfrac{\ln^2(n)}{n^3} \right)=\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$

• $$\left|\dfrac{\dfrac{(-1)^{n}\ln(n)}{n^2}}{\dfrac{\ln(n)}{n^{2}}} \right|\leq 1 \implies \dfrac{(-1)^{n}\ln(n)}{n^2}=\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$
• $$\left|\dfrac{\dfrac{\ln^2(n)}{n^3} }{\dfrac{\ln(n)}{n^{2}}} \right|=\dfrac{\ln(n)}{n} \underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow}0 \implies \dfrac{\ln^2(n)}{n^3} =\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$
• Is my proof correct

Answer 1

Correct, but there are unnecessary details. A shorter version is this $$\dfrac{(-1)^{n}}{n-\ln(n)}=\dfrac{(-1)^{n}}{n}\cdot\dfrac1{1-\cfrac{\ln n}n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$ Now $\;\dfrac1{1-u}=1+\mathcal O(u)$, so $$\dfrac{(-1)^{n}}{n-\ln(n)}=\dfrac{(-1)^{n}}{n}\biggl(1+\mathcal{O}\Bigl(\dfrac{\ln n}{n} \Bigr)\biggr)=\dfrac{(-1)^{n}}{n}+\dfrac{(-1)^{n}}{n}\mathcal{O}\Bigl(\dfrac{\ln n}{n} \Bigr)=\dfrac{(-1)^{n}}{n}+\mathcal{O}\Bigl(\dfrac{\ln n}{n^2}\Bigr).$$