每日一题[638]裂项放缩

已知$n\in\mathcal N^*$，求证：$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots +\dfrac{1}{3n+1}<\dfrac{11}{10}$．

分析与解　设待证不等式左边为$S_n$，则$$S_{n+1}-S_n=\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4}-\dfrac{1}{n+1}=\dfrac{2}{(3n+2)(3n+3)(3n+4)},$$于是原不等式等价于证明$$\dfrac{2}{5\cdot 6\cdot 7}+\dfrac{2}{8\cdot 9\cdot 10}+\cdots +\dfrac{2}{(3n+2)(3n+3)(3n+4)}<\dfrac{11}{10}-S_1,$$也即$$T_n=\dfrac{1}{5\cdot 6\cdot 7}+\dfrac{1}{8\cdot 9\cdot 10}+\cdots +\dfrac{1}{(3n+2)(3n+3)(3n+4)}<\dfrac{1}{120},$$考虑通项\[\begin{split} \dfrac{1}{(3n+2)(3n+3)(3n+4)}&=\dfrac{1}{27}\cdot \dfrac{1}{\left(n+\dfrac 23\right)(n+1)\left(n+\dfrac 43\right)}\\&<\dfrac{1}{27}\cdot \dfrac{1}{n(n+1)(n+2)}\\&=\dfrac{1}{54}\cdot \left[\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right],\end{split} \]后移放缩起点，有$$T_n<\dfrac{1}{5\cdot 6\cdot 7}+\dfrac{1}{54}\cdot \dfrac{1}{6}<\dfrac{1}{120},$$因此原不等式得证．

思考与总结　如果利用积分放缩，可得$S_n<\ln 3+\dfrac{1}{3n+1}$，此时对$n$有要求(事实上为$n\geqslant 240$)．