上册 1.1 数列极限 第17题

\section*{解题过程：} （1）由于 $$ \sum_{k=1}^{n} \frac{1}{n+1} f\left(\frac{k}{n}\right) \leqslant \frac{1}{n+1} f\left(\frac{1}{n}\right)+\frac{1}{n+\frac{1}{2}} f\left(\frac{2}{n}\right)+\cdots+\frac{1}{n+\frac{1}{n}} f\left(\frac{n}{n}\right) \leqslant \sum_{k=1}^{n} \frac{1}{n+\frac{1}{n}} f\left(\frac{k}{n}\right) $$ 且 $$ \begin{aligned} & \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{n+1} f\left(\frac{k}{n}\right)=\lim _{n \rightarrow \infty} \frac{n}{n+1} \sum_{k=1}^{n} \frac{1}{n} f\left(\frac{k}{n}\right)=\int_{0}^{1} f(x) \mathrm{d} x \\ & \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{n+\frac{1}{n}} f\left(\frac{k}{n}\right)=\lim _{n \rightarrow \infty} \frac{n}{n+\frac{1}{n}} \sum_{k=1}^{n} \frac{1}{n} f\left(\frac{k}{n}\right)=\int_{0}^{1} f(x) \mathrm{d} x \end{aligned} $$ 所以 $\displaystyle \lim _{n \rightarrow \infty}\left(\frac{1}{n+1} f\left(\frac{1}{n}\right)+\frac{1}{n+\frac{1}{2}} f\left(\frac{2}{n}\right)+\cdots+\frac{1}{n+\frac{1}{n}} f\left(\frac{n}{n}\right)\right)=\int_{0}^{1} f(x) \mathrm{d} x$ ． 于是（1） $\displaystyle \lim _{n \rightarrow \infty}\left(\frac{\sin \frac{\pi}{n}}{n+1}+\frac{\sin \frac{2 \pi}{n}}{n+\frac{1}{2}}+\cdots+\frac{\sin \frac{n \pi}{n}}{n+\frac{1}{n}}\right)=\int_{0}^{1} \sin \pi x \mathrm{~d} x=\frac{2}{\pi}$ ； （2） $\displaystyle \lim _{n \rightarrow \infty}\left(\frac{a^{\frac{1}{n}}}{n+1}+\frac{a^{\frac{2}{n}}}{n+\frac{1}{2}}+\cdots+\frac{a^{\frac{n}{n}}}{n+\frac{1}{n}}\right)=\int_{0}^{1} a^{x} \mathrm{~d} x=\frac{a-1}{\ln a}$ ． （2）$\displaystyle \sum_{k=1}^{n} \frac{a^{\frac{k}{n}}}{n+(a-1) k^{-1}}=\sum_{k=1}^{n} \frac{1}{n} a^{\frac{k}{n}} \frac{1}{1+(a-1) \frac{1}{n k}}$ ． 由于 $\displaystyle \frac{1}{1+(a-1) \frac{1}{n}} \sum_{k=1}^{n} \frac{1}{n} a^{\frac{k}{n}}<\sum_{k=1}^{n} \frac{a^{\frac{k}{n}}}{n+(a-1) k^{-1}}<\sum_{k=1}^{n} \frac{1}{n} a^{\frac{k}{n}}$ ， 且 $\displaystyle \quad \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{n} a^{\frac{k}{n}}=\int_{0}^{1} a^{x} \mathrm{~d} x=\frac{1}{\ln a}(a-1)$ ， $$ \lim _{n \rightarrow \infty} \frac{1}{1+(a-1) \frac{1}{n}} \sum_{k=1}^{n} \frac{1}{n} a^{\frac{k}{n}}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{n} a^{\frac{k}{n}}=\int_{0}^{1} a^{x} \mathrm{~d} x=\frac{1}{\ln a}(a-1) $$ 所以 $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{a^{\frac{k}{n}}}{n+(a-1) k^{-1}}=\frac{1}{\ln a}(a-1) $$ （3）由中值定理 $\displaystyle \sqrt{1+\frac{k}{n^{2}}}-1=\frac{1}{2} \frac{1}{\sqrt{1+\theta \frac{k}{n^{2}}}} \frac{k}{n^{2}}, \theta \in(0,1)$ ．于是 $\displaystyle 1 \leqslant \sqrt{1+\theta \frac{k}{n^{2}}} \leqslant \sqrt{1+\frac{1}{n}}$ ．从而 $$ \frac{1}{2} \frac{1}{\sqrt{1+\frac{1}{n}}} \frac{k}{n^{2}} \leqslant \sqrt{1+\frac{k}{n^{2}}}-1 \leqslant \frac{1}{2} \frac{k}{n^{2}} $$ 所以 $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\sqrt{1+\frac{k}{n^{2}}}-1\right)=\lim _{n \rightarrow \infty} \frac{1}{2} \sum_{k=1}^{n} \frac{k}{n^{2}}=\frac{1}{2} \int_{0}^{1} x \mathrm{~d} x=\frac{1}{4} $$ （4）类似于（3）有 $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\sqrt{1+\frac{k^{2}}{n^{3}}}-1\right)=\frac{1}{2} \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{2}}{n^{3}}=\frac{1}{2} \int_{0}^{1} x^{2} \mathrm{~d} x=\frac{1}{6}$ ． （5）类似于（3）有 $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\mathrm{e}^{\frac{1}{n} \sin \frac{k}{n}}-1\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sin \frac{k}{n}=\int_{0}^{1} \sin x \mathrm{~d} x=1-\cos 1$ ． （6） $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{\sqrt{k(n-k+1)}}=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k-1}{n}\right)}}=\int_{0}^{1} \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x=\pi$ ． （7） $\displaystyle \lim _{n \rightarrow \infty} \frac{1}{n^{2}} \sum_{k=1}^{n} \sqrt{(n x+k)(n x+k-1)}$ $$ =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \sqrt{\left(x+\frac{k}{n}\right)\left(x+\frac{k-1}{n}\right)}=\int_{0}^{1} \sqrt{(x+t)^{2}} \mathrm{~d} t=\int_{0}^{1}(x+t) \mathrm{d} t=x+\frac{1}{2} . $$