Algebra: Prove inequality $\sum_{n=1}^{2015} \frac1{n^3} < \frac 54$ Can someone prove inequality (n is natural):$$\sum_{n=1}^{2015} \frac{1}{n^3} < \frac 5 4$$
I have tried some predictions like $a^3 > a(a - 1)(a - 2) $ but couldn't get anything out of them.

Note that for $n>1$,
$$\frac{1}{n^3}<\frac{1}{n^3-n}=\frac{1}{2(n-1)}-\frac{1}{n}+\frac{1}{2(n+1)}$$
$$\sum_{n=2}^{2015}\frac{1}{n^3-1}=\frac{1}{2(1)}-\frac{1}{2(2)}-\frac{1}{2(2015)}+\frac{1}{2(2016)}=\frac{2031119}{8124480}<\frac{1}{4}$$
