SPOILER:
On a:
$ S_n=\sum_{1}^{n-1}\sin (\frac{k\pi}{n})=\sum_{0}^{n-1}Im(e^{i\frac{k\pi}{n}}) $
Soit $ S_n=Im(\frac{1-e^{\frac{i\pi}{n}^{n}}}{1-e^{\frac{i\pi}{n}}}) $
$ S_n=Im(\frac{2}{1-e^{i\frac{\pi}{n}}})=Im(\frac{2}{1-e^{i\frac{\pi}{n}}}*\frac{1-e^{-i\frac{\pi}{n}}}{1-e^{-i\frac{\pi}{n}}})=Im(\frac{2(1-e^{-i\frac{\pi}{n}})}{1-e^{i\frac{\pi}{n}}-e^{-i\frac{\pi}{n}}+1} $$ =Im(\frac{2(1-e^{-i\frac{\pi}{n}})}{2-2cos(\frac{\pi}{n})})=Im(\frac{1-e^{-i\frac{\pi}{n}}}{1-1cos(\frac{\pi}{n})})=\frac{-sin(\frac{-\pi}{n})}{1-cos(\frac{\pi}{n})} $
$ S_n=\frac{sin(\frac{\pi}{n})}{1-cos(\frac{\pi}{n})}=\frac{2sin\frac{\pi}{2n}cos\frac{\pi}{2n}}{1-(1-2sin^{2}(\frac{\pi}{2n}))} $
$ S_n=\frac{cos\frac{\pi}{2n}}{sin(\frac{\pi}{2n})}=\frac{1}{tan(\frac{\pi}{2n})} $
Soit$ \lim_{x\rightarrow \infty}\frac{(S_n)}{n}=\lim_{x\rightarrow \infty}(\frac{n}{tan(\frac{\pi}{2n})})=\infty $
$ S_n=\sum_{1}^{n-1}\sin (\frac{k\pi}{n})=\sum_{0}^{n-1}Im(e^{i\frac{k\pi}{n}}) $
Soit $ S_n=Im(\frac{1-e^{\frac{i\pi}{n}^{n}}}{1-e^{\frac{i\pi}{n}}}) $
$ S_n=Im(\frac{2}{1-e^{i\frac{\pi}{n}}})=Im(\frac{2}{1-e^{i\frac{\pi}{n}}}*\frac{1-e^{-i\frac{\pi}{n}}}{1-e^{-i\frac{\pi}{n}}})=Im(\frac{2(1-e^{-i\frac{\pi}{n}})}{1-e^{i\frac{\pi}{n}}-e^{-i\frac{\pi}{n}}+1} $$ =Im(\frac{2(1-e^{-i\frac{\pi}{n}})}{2-2cos(\frac{\pi}{n})})=Im(\frac{1-e^{-i\frac{\pi}{n}}}{1-1cos(\frac{\pi}{n})})=\frac{-sin(\frac{-\pi}{n})}{1-cos(\frac{\pi}{n})} $
$ S_n=\frac{sin(\frac{\pi}{n})}{1-cos(\frac{\pi}{n})}=\frac{2sin\frac{\pi}{2n}cos\frac{\pi}{2n}}{1-(1-2sin^{2}(\frac{\pi}{2n}))} $
$ S_n=\frac{cos\frac{\pi}{2n}}{sin(\frac{\pi}{2n})}=\frac{1}{tan(\frac{\pi}{2n})} $
Soit$ \lim_{x\rightarrow \infty}\frac{(S_n)}{n}=\lim_{x\rightarrow \infty}(\frac{n}{tan(\frac{\pi}{2n})})=\infty $