I would like to show that : $$(-1)^{n}\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]=3\dfrac{(-1)^{n}}{n}+\mathcal{O}\left( \dfrac{1}{n^2}\right) $$ by starting from the left side and get the right side
My Proof:
Note that :
$$\ln\left(1+x \right)\underset{ \overset { n \rightarrow +\infty } {} } {=}x+\mathcal{O}\left(x^{2} \right)$$
$$\dfrac{n(n+2)}{n^2-n+1}=1+\dfrac{3n-1}{n^2-n+1} $$
$$\dfrac{3n-1}{n^2-n+1}\underset{ \overset { n \rightarrow +\infty } {} } {\sim}\dfrac{3}{n}\underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow}0$$
\begin{align*} (-1)^{n}\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]&=(-1)^{n}\ln\left[ 1+\dfrac{3n-1}{n^2-n+1} \right] \\ &\underset{ \overset { n \rightarrow +\infty } {} } {\sim} (-1)^{n}\ln\left[ 1+\dfrac{3}{n} \right] \\ &= (-1)^{n}\left(\dfrac{3}{n}+\mathcal{O}\left( \dfrac{1}{n^{2}}\right) \right) \\ &=3\dfrac{(-1)^{n}}{n}+\mathcal{O}\left( \dfrac{1}{n^2}\right) \end{align*}
$$\fbox{$(-1)^{n}\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]=3\dfrac{(-1)^{n}}{n}+\mathcal{O}\left( \dfrac{1}{n^2}\right) $$ $}$$
- Is my proof correct ?
Your proof is correct. But going along your steps, I would rather write $$ \begin{align} \frac{n(n+2)}{n^2-n+1}=\frac{1+\frac2{n}}{1-\frac1n+\frac1{n^2}}=\left(1+\frac2{n}\right)\left(1+\frac1n+O\left(\frac1{n^2}\right)\right)=1+\frac3n+O\left(\frac1{n^2}\right) \end{align} $$ giving $$ \ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]=\frac3n+O\left(\frac1{n^2}\right) $$ and $$ (-1)^n\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]=3\frac{(-1)^{n}}n+O\left(\frac1{n^2}\right). $$ I think it is clearer to avoid using the symbol $\sim$ within your equalities.