一道丑陋的放缩
已知函数\(f(x)=\ln(x+1)-\lambda x+\dfrac{x^2}{2}(x>0)\)
\((1)\) 若\(f(x)>0\)求\(\lambda\)的取值范围
\((2)\)证明：\(2\ln(n+1)-\dfrac{33}{20}<\displaystyle\sum\limits_{i=1}^n\left(\dfrac{2}{i}-\dfrac{1}{i^2}\right)<2\ln(n+1)\)
解
\((1)\) \(f(0)=0,f^{\prime}(x)=\dfrac{1}{x+1}-\lambda+x,f^{\prime}(0)=1-\lambda\)
若\(1-\lambda<0\)，则\(f^{\prime}(0)<0\)
从而由保号性，一定存在\((0,x_0)\)使得\(f^{\prime}(x)<0\)
进而在\(x\in(0,x_0)\)上\(f(x)<f(0)=0\)不合题!
当\(\lambda\leq 1\)时，有\(f^{\prime}(x)=\dfrac{1}{x+1}+x+1-1-\lambda\geq2-1-\lambda=1-\lambda\geq 0\)
从而\(f(x)>f(0)=0\)合题意
综上\(\lambda\leq 1\)
\((2)\) 由\((1)\)知，取\(\lambda=1\)有\(\ln(x+1)-x+\dfrac{x^2}{2}>0\)
即\(2x-x^2<2\ln(x+1)\)
用\(\dfrac{1}{i}\)代换\(x,i=1,2,3\cdots n\)累加有
\(\displaystyle\sum\limits_{i=1}^{n}\left(\dfrac{2}{i}-\dfrac{1}{i^2}\right)<2\ln(n+1)\)
因\(\ln(1+x)\leq x\)，则\(\ln\left(i+1\right)-\ln i< \dfrac{1}{i},I=1,2\cdots n\)
累加有\(\ln(n+1)<\displaystyle\sum\limits_{i=1}^{n}\dfrac{1}{i}\)
考虑\(\displaystyle \sum\limits_{i=1}^{n}\dfrac{1}{i^2}<\displaystyle \sum\limits_{i=1}^{n}\dfrac{1}{i(i-1)}=1+\dfrac{1}{4}+\dfrac{1}{9}+\displaystyle\sum\limits_{i=4}^{n}\left(\dfrac{1}{i}-\dfrac{1}{i-1}\right)=\dfrac{49}{36}+\dfrac{1}{n}-\dfrac{1}{3}=\dfrac{1}{n}+\dfrac{37}{36}<\dfrac{1}{4}+\dfrac{37}{36}=\dfrac{23}{18}<\dfrac{33}{20}\)
所以\(2\displaystyle\sum\limits_{i=1}^{n}\left(\dfrac{1}{i}—\dfrac{1}{i^2}\right)>2\ln(n+1)-\dfrac{33}{20}\)