# 2011年考研数二第19题解析：函数单调性、微分中值定理、定积分、数列

## 题目

$(Ⅰ)$ 证明：对任意的正整数 $n$, 都有 $\frac{1}{n+1} < \ln(1+\frac{1}{n}) < \frac{1}{n}$ 成立.

$(Ⅱ)$ 设 $a_{n} =$ $1+$ $\frac{1}{2}+$ $…$ $+\frac{1}{n}$ $- \ln n$ $(n=1,2,…)$, 证明数列 ${a_{n}}$ 收敛.

## 解析

### 第 $(Ⅰ)$ 问

#### 方法一

$$\frac{1}{n+1} = \frac{1}{\frac{1}{x} + 1} = \frac{x}{x+1};$$

$$\ln(1+\frac{1}{n}) = \ln (1+x);$$

$$\frac{1}{n} = x.$$

$$f(x) = \ln(1+x) – \frac{x}{x+1}; (1)$$

$$g(x) = x – \ln(1+x). (2)$$

$$(1) 式 \Rightarrow$$

$$f^{‘}(x) = \frac{1}{1+x} – \frac{x+1-x}{(x+1)^{2}} \Rightarrow$$

$$f^{‘}(x) = \frac{1}{1+x} – \frac{1}{(x+1)^{2}} \Rightarrow$$

$$f^{‘}(x) = \frac{x}{(1+x)^{2}} \Rightarrow$$

$$\ln(1+x) > \frac{x}{x+1} \Rightarrow$$

$$\ln(1+\frac{1}{n}) > \frac{1}{n+1}.$$

$$(2) 式 \Rightarrow$$

$$g^{‘}(x) = 1-\frac{1}{1+x} \Rightarrow$$

$$x > \ln(1+x) \Rightarrow$$

$$\frac{1}{n} > \ln(1+\frac{1}{n}).$$

#### 方法二

$$f(x) = \ln(1+x).$$

$$f(0) = 0; f^{‘}(x) = \frac{1}{1+x}.$$

$$\frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}-0}=f^{‘}(\varepsilon) \Rightarrow$$

$$\ln(1+\frac{1}{n}) = \frac{1}{n(1+\varepsilon)}. (3)$$

$$\frac{1}{1+\frac{1}{n}} < \frac{1}{1+\varepsilon} < \frac{1}{1+0} \Rightarrow$$

$$\frac{n}{1+n} < \frac{1}{1+\varepsilon} < 1 \Rightarrow$$

$$\frac{n}{1+n} \cdot \frac{1}{n} < \frac{1}{1+\varepsilon} \cdot \frac{1}{n} < 1 \cdot \frac{1}{n} \Rightarrow$$

$$\frac{1}{1+n} < \frac{1}{n(1+\varepsilon)} < \frac{1}{n}.$$

$$\frac{1}{1+n} < \ln(1+\frac{1}{n}) < \frac{1}{n}.$$

#### 方法三

$$\frac{1}{n+1} \leqslant \frac{1}{x} \leqslant \frac{1}{n}. (4)$$

$$\int_{n}^{n+1} \frac{1}{n+1} dx \leqslant \int_{n}^{n+1} \frac{1}{x} dx \leqslant \int_{n}^{n+1} \frac{1}{n} dx.$$

$①:$

$$\int_{n}^{n+1} \frac{1}{n+1} dx =$$

$$\frac{1}{n+1} \int_{n}^{n+1} 1 dx =$$

$$\frac{1}{n+1} \cdot x|_{n}^{n+1} =$$

$$\frac{1}{n+1} \cdot 1 = \frac{1}{n+1}.$$

$②:$

$$\int_{n}^{n+1} \frac{1}{x} dx =$$

$$\ln x |_{n}^{n+1} =$$

$$\ln (n+1) – \ln (n) =$$

$$\ln (\frac{n+1}{n}) = \ln (1 + \frac{1}{n}).$$

$③:$

$$\int_{n}^{n+1} \frac{1}{n} dx =$$

$$\frac{1}{n} \int_{n}^{n+1} 1 dx =$$

$$\frac{1}{n} \cdot x |_{n}^{n+1} =$$

$$\frac{1}{n} \cdot 1 = \frac{1}{n}.$$

### 第 $(Ⅱ)$ 问

$$a_{n}=\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdot \cdot \cdot + \frac{1}{n} – \ln n.$$

$$a_{n+1}=\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdot \cdot \cdot + \frac{1}{n} + \frac{1}{n+1} – \ln (n+1).$$

$$a_{n+1} – a_{n} =$$

$$\frac{1}{n+1} – \ln (n+1) + \ln n =$$

$$\frac{1}{n+1} – [\ln (n+1) – \ln n] =$$

$$\frac{1}{n+1} – \ln (1+\frac{1}{n}).$$

$$\frac{1}{n+1} < \ln (1 + \frac{1}{n}).$$

$$\frac{1}{n+1} – \ln (1+\frac{1}{n}) < 0.$$

$$a_{n+1} – a_{n} < 0.$$

$$a_{n}=\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdot \cdot \cdot + \frac{1}{n} – \ln n.$$

$$\frac{1}{x} > \ln(1+\frac{1}{x}).$$

$$a_{n} > \ln (1+\frac{1}{1}) + \ln (1+\frac{1}{2}) + \cdot \cdot \cdot + \ln (1+\frac{1}{n}) – \ln n \Rightarrow$$

$$a_{n} > \ln (\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \cdot \cdot \cdot \frac{n+1}{n}) – \ln n \Rightarrow$$

$$a_{n} > \ln(n+1) – \ln n > 0.$$