867

已知函数 ${\displaystyle f(x)=\ln (1-x)+a\sin x}$.

1. 若 ${\displaystyle f(x)<0}$ 对任意的 ${\displaystyle x\in (0,1)}$ 恒成立，求 ${\displaystyle a}$ 的取值范围;
2. 证明：${\displaystyle {\displaystyle \frac{1}{2}\cos \frac{1}{2}+\frac{1}{3}\cos \frac{1}{3}+\cdots +\frac{1}{n}\cos \frac{1}{n}<\ln n(n\in {\mathbb{N}}^{\ast },n\ge 2)}}$.

1. 解 当 ${\displaystyle a\le 1}$，${\displaystyle f(x)\le \ln (1-x)+\sin x=g(x)}$，则 ${\displaystyle g^{\prime }(x)={\displaystyle \cos x-\frac{1}{1-x}<\cos x-1<0}}$，则 ${\displaystyle g(x)<g(0)=0}$，满足题意；当 ${\displaystyle a>1}$，${\displaystyle f^{\prime }(x)={\displaystyle a\cos x-\frac{1}{1-x},f^{″}(x)=-a\sin x-\frac{1}{(x-1{)}^2}<0}}$，而 ${\displaystyle f^{\prime }(0),f^{\prime }(1-\frac{1}{a})>0}$，故 ${\displaystyle f(x)}$ 在 ${\displaystyle {\displaystyle (0,1-\frac{1}{a})}}$ 上单增，与题意不符；综上 ${\displaystyle a\le 1}$.
2. 证明 由于$$\begin{array}{r}\ln \frac{n}{n-1}=-\ln \frac{n-1}{n}\ge -\frac{n-1}{n}+1=\frac{1}{n}>\frac{1}{n}\cos \frac{1}{n},\end{array}$$累加即可得证.