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[duaconcuachua98]

Cho dãy số $(u_{n})$ xác định bởi $\left\{\begin{matrix} u_{1}=1 & \\ u_{n+1}=\sqrt{u_{n}\left ( u_{n}+1 \right )\left ( u_{n}+2 \right )\left ( u_{n}+3 \right )+1},\forall n\in \mathbb{N}^{*} & \end{matrix}\right.$

Đặt $S_{n}=\sum_{i=1}^{n}\frac{1}{u_{i}+2}$

Tính $\lim S_{n}$

 

 

[Phuong Thu Quoc]

Cho dãy số $(u_{n})$ xác định bởi $\left\{\begin{matrix} u_{1}=1 & \\ u_{n+1}=\sqrt{u_{n}\left ( u_{n}+1 \right )\left ( u_{n}+2 \right )\left ( u_{n}+3 \right )+1},\forall n\in \mathbb{N}^{*} & \end{matrix}\right.$

Đặt $S_{n}=\sum_{i=1}^{n}\frac{1}{u_{i}+2}$

Tính $\lim S_{n}$

Có $x_{n+1}=\sqrt{\left ( x_{n}^{2}+3x_{n} \right )\left ( x_{n}^{2}+3x_{n}+2 \right )+1}=x_{n}^{2}+3x_{n}+1$                                             (1)

$x_{n+1}-x_{n}=\left ( x_{n}+1 \right )^{2}\geq 0\Rightarrow \left ( x_{n} \right )$ là dãy tăng.

Giả sử $limx_{n}=L\Rightarrow L=L^{2}+3L+1\Rightarrow L=-1$ vô lí

Vậy dãy không bị chặn trên $\Rightarrow limx_{n}=+\infty$

Từ (1) $\Rightarrow x_{n+1}+1=\left ( x_{n} +1\right )\left ( x_{n} +2\right )\Rightarrow \frac{1}{x_{n+1}+1}=\frac{1}{x_{n}+1}-\frac{1}{x_{n}+2}\Rightarrow \frac{1}{x_{n}+2}=\frac{1}{x_{n}+1}- \frac{1}{x_{n+1}+1}$

Vậy $\sum_{i=1}^{n}\frac{1}{x_{i}+2}=\sum_{i=1}^{n}\left ( \frac{1}{x_{i}+1}-\frac{1}{x_{i+1}+1} \right )=\frac{1}{x_{1}+1}-\frac{1}{x_{n+1}+1}$

$\Rightarrow limS_{n}=\frac{1}{2}$