[再寄小读者之数学篇](2014-12-04 $\left(1+\frac{1}{x}\right)^x>\frac{2ex}{2x+1},\forall\ x>0.$)

试证: $$\bex \left(1+\frac{1}{x}\right)^x>\frac{2ex}{2x+1},\forall\ x>0. \eex$$

 

证明 (from Hansschwarzkopf): 对任何$x>0$, 有 \[x\ln\left(1+\frac{1}{x}\right)=x\ln\frac{1+\frac{1}{2x+1}}{1-\frac{1}{2x+1}} =2x\left(\frac{1}{2x+1}+\frac{1}{3(2x+1)^3}+\ldots\right)>\frac{2x}{2x+1} >\ln \frac{2ex}{2x+1},\] 故 \[\left(1+\frac{1}{x}\right)^x>\frac{2ex}{2x+1},\forall\ x>0.\]