Dãy số (un) nào sau đây có giới hạn khác số 1 khi n dần đến vô cùng?

+) Đáp án C:           

Ta có: $u_{n+1}-1=\frac{1}{2}\left(u_n-1\right)$

$\begin{array}{l}{u}_n-1=\frac{1}{2}\left(u_{n-1}-1\right)\\ u_{n-1}-1=\frac{1}{2}\left(u_{n-2}-1\right)\\ u_{n-2}-1=\frac{1}{2}\left(u_{n-3}-1\right)\\ \mathrm{...}\\ u_2-1=\frac{1}{2}\left(u_1-1\right)\end{array}$

$\Rightarrow \left(u_n-1\right)\left(u_{n-1}-1\right)\left(u_{n-2}-1\right)\mathrm{...}\left(u_2-1\right)$

$=\frac{1}{2}\left(u_{n-1}-1\right).\frac{1}{2}\left(u_{n-2}-1\right).\frac{1}{2}\left(u_{n-3}-1\right)\mathrm{...}\frac{1}{2}\left(u_1-1\right)$

$=\frac{1}{2^{n-1}}\left(u_{n-1}-1\right)\left(u_{n-2}-1\right)\mathrm{...}\left(u_2-1\right)\left(u_1-1\right)$

$\Rightarrow u_n-1=\frac{1}{2^{n-1}}\left(u_1-1\right)$

$\Rightarrow u_n=\frac{2016}{2^{n-1}}+1\iff u_n=4032.{\left(\frac{1}{2}\right)}^n+1\Rightarrow \underset{}{\lim }{u}_n=1$