上册 1.1 数列极限 第13题

\section*{解题过程：} （1）由于 $$ \begin{aligned} & \frac{1+2+\cdots+n}{n^{2}+n+n} \leqslant \frac{1}{n^{2}+n+1}+\frac{2}{n^{2}+n+2}+\cdots+\frac{n}{n^{2}+n+n} \leqslant \frac{1+2+\cdots+n}{n^{2}+n+1} \\ & \lim _{n \rightarrow \infty} \frac{1+2+\cdots+n}{n^{2}+n+n}=\lim _{n \rightarrow \infty} \frac{1}{2} \frac{n(n+1)}{n^{2}+n+n}=\frac{1}{2}, \lim _{n \rightarrow \infty} \frac{1+2+\cdots+n}{n^{2}+n+1}=\lim _{n \rightarrow \infty} \frac{1}{2} \frac{n(n+1)}{n^{2}+n+1}=\frac{1}{2}, \end{aligned} $$ 故 $$ \lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}+n+1}+\frac{2}{n^{2}+n+2}+\cdots+\frac{n}{n^{2}+n+n}\right)=\frac{1}{2} $$ （2）由于 $$ \frac{1}{n^{2}+2 n-1} \cdot \frac{1+(2 n-1)}{2} n \leqslant \frac{1}{n^{2}+1}+\frac{3}{n^{2}+3}+\cdots+\frac{2 n-1}{n^{2}+2 n-1} \leqslant \frac{1}{n^{2}+1} \cdot \frac{1+(2 n-1)}{2} n, $$ $$ \begin{aligned} & \lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}+2 n-1} \cdot \frac{1+2 n-1}{2} n\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}+1} \cdot \frac{1+2 n-1}{2} n\right)=1, \\ & \lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}+1}+\frac{3}{n^{2}+3}+\cdots+\frac{2 n-1}{n^{2}+2 n-1}\right)=1 . \end{aligned} $$ 故 $$ \begin{aligned} & \frac{n}{\sqrt{n^{2}+n}} \leqslant \frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+\cdots+\frac{1}{\sqrt{n^{2}+n}} \leqslant \frac{n}{\sqrt{n^{2}+1}} \\ & \lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}+1}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n^{2}}}}=1, \lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}+n}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n}}}=1 \end{aligned} $$ 故 $$ \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+\cdots+\frac{1}{\sqrt{n^{2}+n}}\right)=1 $$ （4）由于 且 $$ \begin{aligned} & \frac{n}{\sqrt{n^{2}-1}} \leqslant \frac{1}{\sqrt{n^{2}-1}}+\frac{1}{\sqrt{n^{2}-2}}+\cdots+\frac{1}{\sqrt{n^{2}-n}} \leqslant \frac{n}{\sqrt{n^{2}-n}} \\ & \lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}-1}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1-\frac{1}{n^{2}}}}=1, \lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}-n}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1-\frac{1}{n}}}=1 \end{aligned} $$ 故 $$ \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^{2}-1}}+\frac{1}{\sqrt{n^{2}-2}}+\cdots+\frac{1}{\sqrt{n^{2}-n}}\right)=1 . $$ （5）由于 $$ \frac{n}{n+\sqrt{n}} \leqslant \frac{1}{n+1}+\frac{1}{n+\sqrt{2}}+\frac{1}{n+\sqrt{3}}+\cdots+\frac{1}{n+\sqrt{n}} \leqslant \frac{n}{n+1} $$ 且 $$ \lim _{n \rightarrow \infty} \frac{n}{n+\sqrt{n}}=\lim _{n \rightarrow \infty} \frac{n}{n+1}=1 $$ 故 $$ \lim _{n \rightarrow \infty}\left(\frac{1}{n+1}+\frac{1}{n+\sqrt{2}}+\frac{1}{n+\sqrt{3}}+\cdots+\frac{1}{n+\sqrt{n}}\right)=1 $$ （6）方法 1：由于 $$ 0 \leqslant \frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}+\cdots+\frac{1}{(2 n)^{2}} \leqslant \underbrace{\frac{1}{n^{2}}+\frac{1}{n^{2}}+\cdots+\frac{1}{n^{2}}}_{n \uparrow}=\frac{1}{n} \text {, 且 } \lim _{n \rightarrow \infty} \frac{1}{n}=0 \text {, } $$ 故 $$ \lim _{n \rightarrow \infty}\left[\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}+\cdots+\frac{1}{(2 n)^{2}}\right]=0 $$ 方法 2：由级数 $\displaystyle \sum \frac{1}{n^{2}}$ 收玫及柯西收玫准则得证。 （7）由于 $$ \frac{2 n+2}{n+1} \leqslant \sum_{k=n^{2}}^{(n+1)^{2}} \frac{1}{\sqrt{k}} \leqslant \frac{2 n+2}{n} \text {, 且 } \lim _{n \rightarrow \infty} \frac{2 n+2}{n}=\lim _{n \rightarrow \infty} \frac{2 n+2}{n+1}=2 $$ 由两边夹定理可得 $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=n^{2}}^{(n+1)^{2}} \frac{1}{\sqrt{k}}=2$ ． （8）记 $\displaystyle x_{n}=\frac{1}{n^{2}+1}+\frac{2}{n^{2}+\frac{1}{2}}+\cdots+\frac{n}{n^{2}+\frac{1}{n}}$ ，则 $\displaystyle \frac{1+2+\cdots+n}{n^{2}+1} \leqslant x_{n} \leqslant \frac{1+2+\cdots+n}{n^{2}+\frac{1}{n}}$ ．又 $$ \lim _{n \rightarrow \infty} \frac{1+2+\cdots+n}{n^{2}+1}=\lim _{n \rightarrow \infty} \frac{1}{2} \frac{(1+n) n}{\left(n^{2}+1\right)}=\frac{1}{2}, \lim _{n \rightarrow \infty} \frac{1+2+\cdots+n}{n^{2}+\frac{1}{n}}=\frac{1}{2} \lim _{n \rightarrow \infty} \frac{(1+n) n}{n^{2}+\frac{1}{n}}=\frac{1}{2} $$ 故 $$ \lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}+1}+\frac{2}{n^{2}+\frac{1}{2}}+\cdots+\frac{n}{n^{2}+\frac{1}{n}}\right)=\frac{1}{2} $$ （9）由于 $$ \sum_{k=1}^{n} \frac{k^{2}}{n^{3}+2 n+n} \leqslant \sum_{k=1}^{n} \frac{k^{2}}{n^{3}+2 n+k} \leqslant \sum_{k=1}^{n} \frac{k^{2}}{n^{3}+2 n+1} $$ 且 $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{2}}{n^{3}+2 n+n}=\lim _{n \rightarrow \infty} \frac{(n+1)^{2}}{(n+1)^{3}-n^{3}+3}=\frac{1}{3}, \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{2}}{n^{3}+2 n+1}=\lim _{n \rightarrow \infty} \frac{(n+1)^{2}}{(n+1)^{3}-n^{3}+2}=\frac{1}{3}$ ， 故 $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{2}}{n^{3}+2 n+k}=\frac{1}{3} $$ （10）由于 $$ \sqrt{7} \frac{n^{2}}{2 n^{2}+n} \leqslant \frac{\sqrt{7} n}{2 n^{2}+1}+\frac{\sqrt{7} n}{2 n^{2}+2}+\cdots+\frac{\sqrt{7} n}{2 n^{2}+n} \leqslant \sqrt{7} \frac{n^{2}}{2 n^{2}+1} $$ 且 $$ \lim _{n \rightarrow \infty} \sqrt{7} \frac{n^{2}}{2 n^{2}+n}=\lim _{n \rightarrow \infty} \sqrt{7} \frac{n^{2}}{2 n^{2}+1}=\frac{1}{2} \sqrt{7} $$ 故 $$ \lim _{n \rightarrow \infty}\left(\frac{\sqrt{7} n}{2 n^{2}+1}+\frac{\sqrt{7} n}{2 n^{2}+2}+\cdots+\frac{\sqrt{7} n}{2 n^{2}+n}\right)=\frac{1}{2} \sqrt{7} $$ 注：$n$ 项和极限问题一般用夹逼定理、定积分及级数等方法解决．