已知数列$\{x_n\}$满足$x_{n+1}=x_n-\ln x_n$，且$x_1={\rm e}$，求证：$\displaystyle \sum_{k=1}^n\dfrac{x_k-x_{k+1}}{x_k\sqrt{x_k}}<1$．
证明 因为$x-\ln x>1$，所以$x_n>1$恒成立，于是有$x_{n+1}-x_n=-\ln x_n<0$，所以数列$\{x_n\}$单调递减，且$x_n>1$．
方法一 根据题意，有\[\begin{split}\sum_{k=1}^n\dfrac{x_k-x_{k+1}}{x_k\sqrt{x_k}}&<\sum_{k=1}^n\dfrac{x_k-x_{k+1}}{x_k\sqrt{x_{k+1}}}\\&<\sum_{k=1}^n\dfrac{2(x_k-x_{k+1})}{x_k\sqrt{x_{k+1}}+x_{k+1}\sqrt{x_k}}\\&=2\sum_{k=1}^n\left(\dfrac{1}{\sqrt{x_{k+1}}}-\dfrac{1}{\sqrt{x_k}}\right)\\&<2\left(1-\dfrac{1}{\sqrt{\rm e}}\right)<\dfrac 45.\end{split}\]
方法二 设$f(x)=x-\ln x$，考虑$f(x)-\sqrt x=x-\ln x-\sqrt x$，换元后等价于函数$g(t)=t^2-2\ln t-t$，而\[g'(t)=\dfrac{2t^2-t-2}{t},\]于是函数$g(t)$在$\left(0,\dfrac {1+\sqrt{17}}{4}\right]$上单调递减，在$\left(\dfrac{1+\sqrt{17}}{4},\sqrt{\rm e}\right]$上单调递增．又\[g(1)=0,g(\sqrt 2)=2-\ln 2-\sqrt 2<0,\]于是当$x\in (1,2)$时，有$f(x)<\sqrt x$．因此\[\begin{split}\sum_{k=1}^n\dfrac{x_k-x_{k+1}}{x_k\sqrt{x_k}}&<\dfrac{x_1-x_2}{x_1\sqrt{x_1}}+\sum_{k=2}^{n}\dfrac{x_k-x_{k+1}}{x_kx_{k+1}}\\&=\dfrac{x_1-x_2}{x_1\sqrt{x_1}}+\sum_{k=2}^n\left(\dfrac{1}{x_{k+1}}-\dfrac{1}{x_k}\right)\\
&<\dfrac{1}{{\rm e}\sqrt{\rm e}}+1-\dfrac{1}{{\rm e}-1}<\dfrac 23.\end{split}\]
方法三 根据题意，有\[\begin{split}\sum_{k=1}^n\dfrac{x_k-x_{k+1}}{x_k\sqrt{x_k}}&<\sum_{k=1}^n\dfrac{x_k-x_{k+1}}{x_k\sqrt{x_{k+1}}}\\&<\sum_{k=1}^n\dfrac{x_k-x_{k+1}}{x_k}\\&<\sum_{k=1}^n\dfrac{x_k+x_{k+1}}{2x_k}\cdot \left(\ln x_k-\ln x_{k+1}\right)\\&<\sum_{k=1}^{n}\left(\ln x_k-\ln x_{k+1}\right)\\&<1.\end{split}\]
方法四 根据题意，有\[S_n=\sum_{k=1}^n\dfrac{x_k-x_{k+1}}{x_k\sqrt{x_k}}=\sum_{k=1}^n\dfrac{\ln x_k}{x_k^{\frac 32}}<{\rm e}^{-\frac 32}+\dfrac{\ln({\rm e}-1)}{({\rm e}-1)^{\frac 32}}+\sum_{k=3}^n\left(x_k-1\right).\]
另一方面，有\[x_{n+1}-1=x_n-1-\ln x_n,\]于是\[\dfrac{x_{n+1}-1}{x_n-1}=1-\dfrac{\ln x_n}{x_n-1},\]容易证明当$n\geqslant 2$时，有$1<x_n\leqslant {\rm e}-1$，于是\[\dfrac{x_{n+1}-1}{x_n-1}\leqslant 1-\dfrac{\ln({\rm e}-1)}{{\rm e}-2}<\dfrac 14,\]用到了函数$y=\dfrac {\ln x}{x-1}$在$(1,\mathrm e-1)$上单调递减．于是当$n\geqslant 3$时，有\[x_n\leqslant 1+\dfrac 1{
4^{n-3}}\cdot \left({\rm e}-2-\ln({\rm e}-1)\right),\]于是\[S_n<{\rm e}^{-\frac 32}+\dfrac{\ln({\rm e}-1)}{({\rm e}-1)^{\frac 32}}+\dfrac{{\rm e}-2-\ln({\rm e}-1)}{1-\dfrac 14}={\rm e}^{-\frac 32}+\dfrac{\ln({\rm e}-1)}{({\rm e}-1)^{\frac 32}}+\dfrac 43\left({\rm e}-2-\ln({\rm e}-1)\right)<1.\]