(I) 设 $f(x)$ 是 $[0,+\infty)$ 上单调减少且非负的连续函数. 证明:
$$f(k+1) \leqslant \int_k^{k+1} f(x) \mathrm{d} x \leqslant f(k)(k=1,2, \cdots)$$
(II) 证明 : $\ln (1+n) \leqslant 1+\frac{1}{2}+\cdots+\frac{1}{n} \leqslant 1+\ln n$, 并求极限 $\lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{\ln n}$.

$$\int_k^{k+1} f(k+1) \mathrm{d} x \leqslant \int_k^{k+1} f(x) \mathrm{d} x \leqslant \int_k^{k+1} f(k) \mathrm{d} x,$$

(II) 若取 $f(x)=\frac{1}{x}(x>0)$, 显然 $f(x)$ 单调减少且非负, 则由 (I) 知
$$\frac{1}{n+1} \leqslant \int_n^{n+1} \frac{1}{x} \mathrm{~d} x \leqslant \frac{1}{n}$$

\begin{aligned} \int_1^{n+1} \frac{1}{x} \mathrm{~d} x & =\int_1^2 \frac{1}{x} \mathrm{~d} x+\int_2^3 \frac{1}{x} \mathrm{~d} x+\cdots+\int_n^{n+1} \frac{1}{x} \mathrm{~d} x \\ & \leqslant 1+\frac{1}{2}+\cdots+\frac{1}{n}, \\ 1+\int_1^n \frac{1}{x} \mathrm{~d} x & =1+\int_1^2 \frac{1}{x} \mathrm{~d} x+\int_2^3 \frac{1}{x} \mathrm{~d} x+\cdots+\int_{n-1}^n \frac{1}{x} \mathrm{~d} x \\ & \geqslant 1+\frac{1}{2}+\cdots+\frac{1}{n} . \end{aligned}

$$\int_1^{n+1} \frac{1}{x} \mathrm{~d} x=\left.\ln x\right|_1 ^{n+1}=\ln (1+n), \int_1^n \frac{1}{x} \mathrm{~d} x=\left.\ln x\right|_1 ^n=\ln n,$$

$$\ln (1+n) \leqslant 1+\frac{1}{2}+\cdots+\frac{1}{n} \leqslant 1+\ln n$$

$$\frac{\ln (1+n)}{\ln n} \leqslant \frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{\ln n} \leqslant \frac{1+\ln n}{\ln n}$$

$$\begin{gathered} \lim _{n \rightarrow \infty} \frac{1+\ln n}{\ln n}=\lim _{n \rightarrow \infty} \frac{1}{\ln n}+1=1, \\ \lim _{n \rightarrow \infty} \frac{\ln (1+n)}{\ln n}=\lim _{x \rightarrow+\infty} \frac{\ln (1+x)}{\ln x} \stackrel{\text { 洛 }}{=} \lim _{x \rightarrow+\infty} \frac{\frac{1}{1+x}}{\frac{1}{x}}=1, \end{gathered}$$

$$\lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{\ln n}=1$$