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

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

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