Cho dãy $(u_{n})$ thoả mãn: $\left\{\begin{matrix} u_{0}=\frac{1}{2}\\u_{k+1}=u_{k}+\frac{1}{n}u_{k}^{2},\forall k=\overline{0,n-1} \end{matrix}\right.$
Tìm $\lim u_{n}$
Lời giải:
$u_{k+1}=u_{k}+\frac{1}{n}u_{k}^{2}\Rightarrow nu_{k+1}=nu_{k}+u_{k}^{2}\Rightarrow (u_{k+1}-u_{k})(n+u_{k})=u_{k}u_{k+1}$
$\Rightarrow \frac{1}{u_{k}}-\frac{1}{u_{k+1}}=\frac{1}{n+u_{k}}$.
$\Rightarrow \sum_{j=0}^{n-1}(\frac{1}{u_{j}}-\frac{1}{u_{j+1}})=\frac{1}{u_{0}}-\frac{1}{u_{n}}=2-\frac{1}{u_{n}}< n.\frac{1}{n}=1\Rightarrow u_{n}< 1$ (1)
$2-\frac{1}{u_{n}}> \frac{n}{n+1}$ (do $u_{n}<1$ ) $u_{n}> \frac{n+1}{n+2}$ (2)
Từ (1) và (2) $\frac{n+1}{n+2}< u_{n} < 1$.
Theo nguyên lý kẹp ta có $\lim_{n\rightarrow +\infty }u_{n}=1$
- namcpnh, chanhquocnghiem và Drago thích