Test 2

$ \sum_{n=1}^{\infty}\sin^{-1}\left( \frac{1}{\sqrt{n+1}} \right) > \sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}} > \lim_{n\rightarrow\infty}\frac{1}{\sqrt{n+1}}\cdot n= \lim_{n\rightarrow\infty}\left( \sqrt{n+1}-\frac{1}{\sqrt{n+1}} \right)=\infty

$ \sin\theta < \theta < \tan\theta

$ \tan^{-1}x < x < \sin^{-1}x