Chủ đề này có 2 trả lời

[nguyenquangtruonghktcute]

Cho $U_{n}$

 $\left\{\begin{matrix} U_{1}=1 & \\ U_{n+1}=1+U_{1}.U_{2}...U_{n} & \end{matrix}\right.$

Đặt $S_{n}=\sum_{n}^{k=1}\frac{1}{U_{k}}$

Tìm lim $S_{n}$

[manhhung2013]

Cho $U_{n}$

 $\left\{\begin{matrix} U_{1}=1 & \\ U_{n+1}=1+U_{1}.U_{2}...U_{n} & \end{matrix}\right.$

Đặt $S_{n}=\sum_{n}^{k=1}\frac{1}{U_{k}}$

Tìm lim $S_{n}$

$\Rightarrow u_{n}^{2}=u_{n}+(u_{n+1}-1)\Rightarrow u_{n}(u_{n}-1)=u_{n+1}-1=>\frac{1}{u_{n}(u_{n}-1)}=\frac{1}{u_{n+1}-1}\Rightarrow \frac{1}{u_{n}-1}-\frac{1}{u_{n}}=\frac{1}{u_{n+1}-1}\Rightarrow \frac{1}{u_{n}-1}-\frac{1}{u_{n+1}-1}=\frac{1}{u_{n}},\forall n\geqslant2$

Truy hồi:

$S_{n}=1+(\frac{1}{u_{2}-1}-\frac{1}{u_{n+1}-1})=\frac{3}{2}-\frac{1}{u_{n+1}-1}$

$\lim_{n\rightarrow +\propto }u_{n}=+\propto \Rightarrow \lim_{n\rightarrow +\propto }S_{n}=\frac{3}{2}$

[nguyenquangtruonghktcute]

$\Rightarrow u_{n}^{2}=u_{n}+(u_{n+1}-1)\Rightarrow u_{n}(u_{n}-1)=u_{n+1}-1=>\frac{1}{u_{n}(u_{n}-1)}=\frac{1}{u_{n+1}-1}\Rightarrow \frac{1}{u_{n}-1}-\frac{1}{u_{n}}=\frac{1}{u_{n+1}-1}\Rightarrow \frac{1}{u_{n}-1}-\frac{1}{u_{n+1}-1}=\frac{1}{u_{n}},\forall n\geqslant2$

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

Bài viết đã được chỉnh sửa nội dung bởi nguyenquangtruonghktcute: 17-05-2017 - 18:04