Show $\cos\left( \pi n^{2}\ln\left(\frac{n}{n-1} \right) \right)=(-1)^{n+1}\frac{\pi}{3n}+\mathcal{O}\left( \frac{1}{n^2}\right) $

Question

I would like to show : $$\cos\left( \pi n^{2}\ln\left(\dfrac{n}{n-1} \right) \right)=(-1)^{n+1}\dfrac{\pi}{3n}+\mathcal{O}\left( \dfrac{1}{n^2}\right) $$ by starting from the left side and get the right side

My attempt:

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

Answer 1

You may write $$ \begin{align*} \cos\left( \pi n^{2}\ln\left(\dfrac{n}{n-1} \right) \right)&=\cos\left( -\pi n^{2}\ln\left(\dfrac{n-1}{n} \right) \right)\\ &=\cos\left( -\pi n^{2}\ln\left(1-\frac1n \right) \right)\\ &=\cos\left( \pi n^{2}\left(\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{3 n^3}+\mathcal{O}\left(\dfrac{1}{n^4} \right) \right) \right) \\&=\cos\left( \pi n+\frac{\pi}2+\frac{\pi}{3 n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) \right) \\&=-\sin\left( \pi n+\frac{\pi}{3 n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) \right) \\&=(-1)^{n+1}\sin\left(\frac{\pi}{3 n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) \right) \\&=(-1)^{n+1}\frac{\pi}{3 n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) \end{align*} $$ as announced.