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) $ 
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*}

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.
