Algebra: Prove inequality $\sum_{n=1}^{2015} \frac1{n^3} < \frac 54$

You can use (for $n \geq 2$) $$\frac{1}{n^3} < \frac{1}{n^2(n-1)} < \frac12\frac{2(n-1) +1}{n^2 (n-1)^2} =\frac{1}{2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2}\right).$$

Write your sum as $$1 + \frac{1}{2^3} + \sum_{2015\geq n\geq 3} \frac{1}{n^3} < \frac{9}{8} + \sum_{n\geq 3} \frac{1}{2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = \frac{9}{8} +\frac{1}{8} = \frac{5}{4}. $$

Split the sum as $1 + 1/8 + \sum_{n=3}^\infty 1/n^3$

You need $\sum_{n=3}^\infty 1/n^3 \lt 1/8$.

$\sum_{n=3}^\infty 1/n^3 \lt \int_{x=2}^\infty 1/x^3 dx = 1/8$

Note that for $n>1$,

$$\frac{1}{n^3}<\frac{1}{n^3-n}=\frac{1}{2(n-1)}-\frac{1}{n}+\frac{1}{2(n+1)}$$

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