How prove this inequality $\sum_{i=1}^{n}\frac{\ln{i}}{i^4}<\frac{1}{14}$ I need to show that
$$\sum_{i=2}^{n}\dfrac{\ln{i}}{i^4}<\dfrac{1}{14}$$
This problem from high school competition, so usage of integrals and infinite serries is forbidden.
My try: let $x\in (n-1,n)$,then
$$\dfrac{ln{x}}{x^4}<\dfrac{\ln{(n-1)}}{(n-1)^4}?$$
and then I can't proceed. 
Thank you 
 A: Since 
$$\frac{d}{dx}\left[\frac{\log x}{x}\right] = \frac{1 - \log x}{x^2} < 0
\quad\text{ when } x > e$$
We have 
$$\frac{\log k}{k} \le \frac{\log 4}{4} = \frac{\log 2}{2}\quad\text{ for any } k \ge 4.$$ As a result,
$$\sum_{k=2}^\infty\frac{\log k}{k^4}
= \sum_{k=2}^\infty\left(\frac{\log k}{k}\right)\frac{1}{k^3}
\le \frac{\log 2}{2^4} + \frac{\log 3}{3^4} + \frac{\log 2}{2}\sum_{k=4}^\infty\frac{1}{k^3}
$$
Notice the last term is bounded by a telescoping series:
$$\sum_{k=4}^\infty\frac{1}{k^3} < \sum_{k=4}^\infty\frac{1}{k^3-k}
= \sum_{k=4}^\infty\frac{1}{(k-1)k(k+1)}
= \frac12 \sum_{k=4}^\infty\left( \frac{1}{(k-1)k} - \frac{1}{k(k+1)}\right)
= \frac12 \frac{1}{(4-1)4} = \frac{1}{24}
$$
We get
$$\sum_{k=2}^\infty\frac{\log k}{k^4} < \frac{\log 2}{16} + \frac{\log 3}{81} + \frac{\log 2}{48} = \frac{\log 2}{12} + \frac{\log 3}{81} \sim 0.071325 < \frac{1}{14}$$
A: With the Riemann zeta function for $x>1$
$$\zeta(x) = \sum_{n=1}^{\infty}\frac{1}{n^x}$$ you get, taking the derivative term-wise 
$$\zeta'(x) = -\sum_{n=1}^{\infty}\frac{\ln{n}}{n^x}.$$
Since all your terms are positive we have
$$\sum_{i=2}^{n}\dfrac{\ln{i}}{i^4}
=\sum_{i=1}^{n}\dfrac{\ln{i}}{i^4}
< -\zeta'(4) = 0.068911265896125379849\dots < \frac{1}{14}
$$
