Growth rate of $f(n) = \frac{1-\cos(\frac{n\pi}{n+1})}{1-\cos(\frac{\pi}{n+1})}$ I encountered this expression when studying the growth of condition numbers on a linear system. I was trying to verify that the order of the growth is $O(n^2)$. However I'm stuck upon proving this. Any hints would be appreciated. Thanks.

Hint
$$1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$$
Then $$\begin{aligned}
f(n) &= \frac{1-\cos(\frac{n\pi}{n+1})}{1-\cos(\frac{\pi}{n+1})}\\
&= \frac{\sin^2\left(\frac{n \pi }{2(n+1)}\right)}{\sin^2\left(\frac{\pi}{2(n+1)}\right)}\\
& \le \frac{1}{\sin^2\left(\frac{\pi}{2(n+1)}\right)}\\
&\simeq \frac{\pi^2}{4}n^2
\end{aligned}$$
