每日一题[1796]裂项放缩

设 $\alpha,\beta$ 为实数,$n$ 为正整数,且 $0\leqslant \beta\leqslant \alpha\leqslant \dfrac{\pi}{4}$,$n>1$.

1、证明:$\dfrac{\tan\alpha-\tan\beta}{1+\tan^{2}\alpha}\leqslant \alpha-\beta$,并判断等号成立的条件.

2.证明:$\displaystyle \sum\limits_{k=1}^{n}\dfrac{1}{n^{2}+k^{2}}<\dfrac{\pi}{4n}$.

解析

1、根据题意,有\[LHS=\dfrac{\cos\alpha}{\cos\beta}\cdot \sin(\alpha-\beta)\leqslant \alpha-\beta,\]命题得证.

2、令 $x_k=\arctan\dfrac kn$($k=0,1,2,\cdots,n$),则\[\sum_{k=1}^n\dfrac1{n^2+k^2}=\dfrac 1n\sum_{k=1}^n\dfrac{\dfrac{k}n-\dfrac{k-1}n}{1+\left(\dfrac kn\right)^2}=\dfrac 1n\sum_{k=1}^n\dfrac{\tan x_k-\tan x_{k-1}}{1+\tan^2x_k}\leqslant \dfrac{\pi}{4n},\]命题得证.

此条目发表在每日一题分类目录,贴了标签。将固定链接加入收藏夹。

发表评论