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$$\frac{d}{dx}\left[\frac{\log x}{x}\right] = \frac{1 - \log x}{x^2} < 0
\quad\text{当 } x > e$$
我们有
$$\frac{\log k}{k} \le \frac{\log 4}{4} = \frac{\log 2}{2}\quad\text{ 对于 } k \ge 4.$$
则有
$$\sum_{k=2}^\infty\frac{\log k}{k^4}
= \sum_{k=2}^\infty\left(\frac{\log k}{k}\right)\frac{1}{k^3}
\le \frac{\log 2}{2^4} + \frac{\log 3}{3^4} + \frac{\log 2}{2}\sum_{k=4}^\infty\frac{1}{k^3}
$$
则
$$\sum_{k=4}^\infty\frac{1}{k^3} < \sum_{k=4}^\infty\frac{1}{k^3-k}
= \sum_{k=4}^\infty\frac{1}{(k-1)k(k+1)}
= \frac12 \sum_{k=4}^\infty\left( \frac{1}{(k-1)k} - \frac{1}{k(k+1)}\right)
= \frac12 \frac{1}{(4-1)4} = \frac{1}{24}
$$
我们得到
$$\sum_{k=2}^\infty\frac{\log k}{k^4} < \frac{\log 2}{16} + \frac{\log 3}{81} + \frac{\log 2}{48} = \frac{\log 2}{12} + \frac{\log 3}{81} \sim 0.071325 < \frac{1}{14}$$
备注:$$\zeta(x) = \sum_{n=1}^{\infty}\frac{1}{n^x},x>1$$
则有
$$\zeta'(x) = -\sum_{n=1}^{\infty}\frac{\ln{n}}{n^x}.$$
所以
$$\sum_{i=2}^{n}\dfrac{\ln{i}}{i^4}
=\sum_{i=1}^{n}\dfrac{\ln{i}}{i^4}
< -\zeta'(4) = 0.068911265896125379849\cdots < \frac{1}{14}
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战巡
发表于 2013-11-9 05:22
