Chứng minh: $\displaystyle 1+\int_{2}^{n+1} {\cfrac{1}{x-1}}\, dx>\sum_{k=1}^{n} {\cfrac{1}{k}}>\int_{1}^{n+1} {\cfrac{1}{x}}\, dx$

Từ đề bài, dễ dàng nhận thấy $n > 1$

Bất đẳng thức phụ:

Xét $f(x) = \dfrac{1}{x} - \ln(x+1) + \ln(x)$

$\Rightarrow f'(x) = - \dfrac{1}{x^3 + x^2} <0\quad \forall x >1$

$\Rightarrow f(x)$ nghịch biến trên $(1;+\infty)$

Lại có: $\lim\limits_{x\to +\infty}f(x) = 0$

$\Rightarrow f(x) >0\quad \forall x >1$

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

Tương tự, ta có:

$\quad f(x) = \ln(x+1) - \ln(x) - \dfrac{1}{x+1} > 0\quad \forall x >1$

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

Do đó ta được:

$\dfrac{1}{x} > \ln(x+1) - \ln(x) > \dfrac{1}{x+1}$

(Hoặc chứng minh bằng vẽ đồ thị)

Theo bất đẳng thức vừa chứng minh, ta có:

\(\begin{array}{l}\quad \dfrac{1}{k} > \ln(k+1) - \ln(k) > \dfrac{1}{k+1}\qquad (k>1)\\
\Leftrightarrow \displaystyle\sum_{k=1}^n\dfrac{1}{k} > \displaystyle\sum_{k=1}^n\left[\ln(k+1) - \ln(k)\right] > \displaystyle\sum_{k=1}^n\dfrac{1}{k+1}\\
\Leftrightarrow \displaystyle\sum_{k=1}^n\dfrac{1}{k} > \ln(n+1)! - \ln n! >\displaystyle\sum_{k=2}^{n+1}\dfrac{1}{k}\\
\Leftrightarrow \displaystyle\sum_{k=1}^n\dfrac{1}{k} > \ln(n+1) ​>\displaystyle\sum_{k=2}^{n+1}\dfrac{1}{k}\\
\Leftrightarrow 1 + \displaystyle\sum_{k=1}^n\dfrac{1}{k} > 1 + \ln(n+1) ​>1 + \displaystyle\sum_{k=2}^{n+1}\dfrac{1}{k}\\
\Leftrightarrow 1 + \displaystyle\sum_{k=1}^n\dfrac{1}{k} > 1 + \ln(n+1) ​> \displaystyle\sum_{k=1}^{n+1}\dfrac{1}{k}\\
\Leftrightarrow \begin{cases}1 + \displaystyle\sum_{k=1}^n\dfrac{1}{k} > 1 + \ln(n+1)\\ 1 + \ln(n+1) ​> \displaystyle\sum_{k=1}^{n+1}\dfrac{1}{k}\end{cases}\qquad (*)\\
Đặt\ N = n + 1\\
(*)\Leftrightarrow \begin{cases}1 + \displaystyle\sum_{k=1}^{N-1}\dfrac{1}{k} > 1 + \ln N\quad (1)\\ 1 + \ln N ​> \displaystyle\sum_{k=1}^{N}\dfrac{1}{k}\qquad \quad(2)\end{cases}\\
(1) \Leftrightarrow \displaystyle\sum_{k=1}^{N-1}\dfrac{1}{k} > \ln N\\
\Leftrightarrow \displaystyle\sum_{k=1}^n\dfrac{1}{k} > \ln(n+1)\\
\Leftrightarrow \displaystyle\sum_{k=1}^m\dfrac{1}{k} > \ln(m+1)\qquad (I)\\
(2) \Leftrightarrow 1 + \ln m > \displaystyle\sum_{k=1}^{m}\dfrac{1}{k}\quad (II)\\
(I),(II)\Rightarrow 1 + \ln m > \displaystyle\sum_{k=1}^{m}\dfrac{1}{k} > \ln(m+1)\\
\Leftrightarrow 1 + \displaystyle\int\limits_2^{m+1}\dfrac{1}{x-1}dx > \displaystyle\sum_{k=1}^{m}\dfrac{1}{k} > \displaystyle\int\limits_1^{m+1}\dfrac{1}{x}dx\\
\Leftrightarrow 1 + \displaystyle\int\limits_2^{n+1}\dfrac{1}{x-1}dx > \displaystyle\sum_{k=1}^{n}\dfrac{1}{k} > \displaystyle\int\limits_1^{n+1}\dfrac{1}{x}dx\\
\end{array}\)