$\Rightarrow u_{n}^{2}=u_{n}+(u_{n+1}-1)\Rightarrow u_{n}(u_{n}-1)=u_{n+1}-1=>\frac{1}{u_{n}(u_{n}-1)}=\frac{1}{u_{n+1}-1}\Rightarrow \frac{1}{u_{n}-1}-\frac{1}{u_{n}}=\frac{1}{u_{n+1}-1}\Rightarrow \frac{1}{u_{n}-1}-\frac{1}{u_{n+1}-1}=\frac{1}{u_{n}},\forall n\geqslant2$
Truy hồi:
$S_{n}=1+(\frac{1}{u_{2}-1}-\frac{1}{u_{n+1}-1})=\frac{3}{2}-\frac{1}{u_{n+1}-1}$
$\lim_{n\rightarrow +\propto }u_{n}=+\propto \Rightarrow \lim_{n\rightarrow +\propto }S_{n}=\frac{3}{2}$
chứng minh $\lim_{n\rightarrow +\propto }u_{n}=+\propto$ giùm mình đi bạn