1. Cho dãy số$(x_{n})$
$x_{1}=1;x_{n+1}=\frac{x_{n}^{24}}{24}+x_{n}$. Tìm $lim\sum_{k=1}^{n}\frac{x_{k}^{23}}{x_{k+1}}$.
2. Cho dãy số$(x_{n})$;$x_{1}=2013;x_{n+1}=\frac{2013x_{n}(1+x_{n}^{2})}{2013x_{n}^{2}-x_{n}+2013}$
$lim\frac{1}{n} \sum_{k=1}^{n}\frac{x_{k}^{2}}{1+x_{k}^{2}}$
$x_{n+1}-x_{n}=\frac{x_{n}^{24}}{24}$ nên $x_{n}$ là dãy tăng
Giả sử dãy có giới hạn hữu hạn là $L$ $\Rightarrow L=0$ vô lí nên dãy có giới hạn vô cực
$x_{n+1}-x_{n}=\frac{x_{n}^{24}}{24}$
$\Rightarrow \frac{24(x_{n+1}-x_{n})}{x_{n}.x_{n+1}}=\frac{x_{n}^{23}}{x_{n+1}}$
$\Rightarrow \frac{x_{n}^{23}}{x_{n+1}}=24(\frac{1}{x_{n}}-\frac{1}{x_{n+1}})$
$\Rightarrow lim\sum_{k=1}^{n}\frac{x_{k}^{23}}{x_{k+1}}$
$=lim\sum_{k=1}^{n}24(\frac{1}{x_{n}}-\frac{1}{x_{n+1}})=lim24(\frac{1}{x_{1}}-\frac{1}{x_{n+1}})=24$