Calculation of a limit+ 
For $\alpha>1$ how can I show that 
  $\lim\limits_{n\to \infty}\dfrac{\pi}{n}\left( \ln \left(\dfrac{\left(α-1\right)^{2}\left(α^{2n}-1\right)}{(α+1)(\alpha-1)}\right)\right)=2\pi\ln(\alpha)$

Indeed,
$\lim\limits_{n\to \infty}\dfrac{\pi}{n}\left( \ln \left(\dfrac{(α-1)^{2}(α^{2n}-1)}{(α+1)(\alpha-1)}\right)\right)=\lim\limits_{n\to \infty}\left(\dfrac{\pi \ln \left(\dfrac{(α-1)(α^{2n}-1)}{α+1}\right)}{n}\right)=$
since $\alpha>1$ then  $\lim\limits_{n\to\infty\:}\alpha^{2n}-1=+\infty$ 
Any help will be appreciated 

1) Use $\log(xy) = \log x + \log y$ and $\log \frac{x}{y} = \log x - \log y$ to simplify everything except the $\log (a^{2n}-1)$ term; all of them tend to 0
2) $\lim_{n \to \infty}\frac{\pi \log (a^{2n}-1)}{n} = \frac{\pi (\log (a^{2n}) + \log (1-\frac{1}{a^{2n}}))}{n} \sim 2 \pi \log a -\frac{\pi}{n a^{2n}} \to_{n} 2 \pi \log a$
