How to determine if $\sum_{n=1}^{\infty}\left (\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$ converges or diverges $$\sum_{n=1}^{\infty} \left(\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$$
Using root test seems not a efficient way since I got stuck without knowing what to do next 
$$\lim_{n\to\infty}\left(\frac{n^2-5n+1}{n^2-4n+2}\right)^{n}$$
Using ratio test is too complicated
$$\left(\frac{(n+1)^2-5(n+1)+1}{(n+1)^2-4(n+1)+2}\right)^{(n+1)^2}=\left(\frac{n^2-3n-3}{n^2-2n-1}\right)^{n^2+2n+1}$$
$$\to\left(\frac{(n+1)^2-5(n+1)+1}{(n+1)^2-4(n+1)+2}*\frac{n^2-4n+2}{n^2-5n+1}\right)^{n^2}*\left(\frac{n^2-3n-3}{n^2-2n-1}\right)^{2n}*\frac{n^2-3n-3}{n^2-2n-1}$$

The general term decays like $\left(1-\frac{1}{n}\right)^{n^2}\approx e^{-n}$ hence the series is convergent.
