试求 $$ex maxsed{al;sex{1+frac{1}{n}}^{n+al}leq e,quad forall ninbN}. eex$$
解答: $$eex ea &quad sex{1+frac{1}{n}}^{n+al}leq e,quadforall ninbN\ &lra (n+al)lnsex{1+frac{1}{n}}leq 1,quadforall ninbN\ &lra alleq frac{1}{lnsex{1+frac{1}{n}}}-n,quadforall ninbN\ &lra alleq inf_{ngeq 1}sed{frac{1}{lnsex{1+frac{1}{n}}}-n} eea eeex$$ 设 $$ex f(x)=frac{1}{ln (1+x)}-frac{1}{x},quad 0<x<1, eex$$ 则 $$ex f'(x)=-frac{1}{(1+x)ln^2(1+x)}+frac{1}{x^2} =frac{-x^2+(1+x)ln^2(1+x)}{x^2(1+x)ln ^2(1+x)}. eex$$ 令 $$ex g(x)=-x^2+(1+x)ln^2(1+x), eex$$ 则 (上课时对不起了, 算了一阶导数就没算下去了...所以一定要坚持算下去...) $$eex ea g'(x)&=-2x+ln^2(1+x)+2ln(1+x),\ g''(x)&=-2+frac{2ln(1+x)}{1+x}+frac{2}{1+x} =frac{2[ln(1+x)-x]}{1+x}leq 0. eea eeex$$ 故 $$eex ea serd{a{rr} serd{a{rr} g''(x)leq 0\ g'(0)=0 ea} a g'(x)leq 0\ g(0)=0 ea} a g(x)leq 0 a f'(x)leq 0; eea eeex$$ $$ex inf_{0<x<1}f(x)=f(1)=frac{1}{ln 2}-1; eex$$ $$ex maxsed{al;sex{1+frac{1}{n}}^{n+al}leq e,quad forall ninbN} =frac{1}{ln 2}-1. eex$$
注记1. 同上讨论, $$ex minsed{al;sex{1+frac{1}{n}}^{n+eta}geq e,quad forall ninbN} =lim_{x o 0} f(x)=frac{1}{2}. eex$$
注记2. 我们也得到了一个更强的对数不等式: $$ex frac{x}{1+x}<ln(1+x)<frac{x}{sqrt{1+x}},quad 0<x<1. eex$$