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这个极限怎么求

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dim Posted 2016-8-1 12:44 |Read mode
Last edited by hbghlyj 2025-5-16 22:41Karamata 不等式:设 $x>0, x \neq 1$,则
\[
\frac{\ln x}{x-1} \leqslant \frac{1+\sqrt[3]{x}}{x+\sqrt[3]{x}}
\]
令 $x=\left(\frac{1+t}{1-t}\right)^3, 0<|t|<1$,并利用 $\ln \frac{1+t}{1-t}$ 的 Taylor 级数展开式,则所证不等式变成下述显然成立的不等式:
\[
3 \sum_{k=0}^{\infty}\left(1-\frac{1}{4 k+1}\right) t^{4 k}+\sum_{k=0}^{\infty}\left(1-\frac{3}{4 k+3}\right) t^{4 k+2} \geqslant 0 .
\]
令 $f(n)=1+\frac{1}{2}\left(\frac{e}{n}\right)^n n!-\frac{n!}{n^n} \sum_{k=0}^n \frac{n^k}{k!}$.利用 Karamata 不等式证明 $\frac{1}{3}<f(n)$ $<\frac{1}{2}$,更确切地,当 $n$ 从 0 增到 $\infty$ 时,$f(n)$ 从 $\frac{1}{2}$ 递减地趋于 $\frac{1}{3}$.

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 Author| dim Posted 2016-8-1 13:11
好吧,其实这个结论比这里的还弱forum.php?mod=viewthread&tid=3992&extra=page=1。。。请问如何证明f(n)递减呢?

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