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

Let $S = \sum_{k=1}^{2015} \frac{1}{k^3}$. Then, since ${\textstyle \frac{1}{x^3}}$ decreases as $x$ increases, we have ${\textstyle \frac{1}{k^3} < \int_{k-1}^{k} \frac{dx}{x^3}}$ for all integers $k >1$. Consequently $S - (\frac{1}{1^3} + \frac{1}{2^3}) = \sum_{k=3}^{2015} \frac{1}{k^3} < \int_{2}^{2015} \frac{dx}{x^3} = \Big[ -\frac{1}{2x^2} \Big]_{2}^{2015} = \frac{1}{2^3} - \frac{1}{2 \cdot 2015^2}$, yielding ${\textstyle S < \frac{5}{4}}$.

$$\frac{1}{2^3} + \frac{1}{3^3} + \dots < \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots = \frac{1}{2} ( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} + \dots) = \frac{1}{4}.$$

Prove inequallity : $$1+\frac{1}{2^4}+\frac{1}{3^4}+...+\frac{1}{n^4}<\frac{11}{10}.$$

$\int_3^{2015} \frac{1}{x^4} < \frac{1}{81}$, so the sum $< 1 + \frac{1}{16} + \frac{1}{81} + \frac{1}{81} = 1 + \frac{113}{1296} < \frac{11}{10}$

@above note that the stopping point is $n$, not $2015$.

Where can I find more inequalities similar to this one?

Induction works pretty well: Suppose that for $n$ we have $$1+\frac{1}{2^3}+...+\frac{1}{n^3} < \frac{5}{4} - \frac{n-1}{n^3}$$Then we want to show that $$1+\frac{1}{2^3}+...+\frac{1}{n^3}+\frac{1}{(n+1)^3} < \frac{5}{4} - \frac{n}{(n+1)^3}=\frac{5}{4}-\frac{n-1}{n^3}+\frac{n-1}{n^3}-\frac{n}{(n+1)^3}$$So if we show that $\frac{1}{(n+1)^3} < \frac{n-1}{n^3}+\frac{n}{(n+1)^3}$, it's done. But this is equivalent to $\frac{1}{(n+1)^2} < \frac{n-1}{n^3}$ $<=>$ $n^3 < n^3+2n^2+n-n^2-2n-1$ $<=>$ $n^2-2n-1>0$, which is true for $n \le 3$. For $n=2$ the sum is equal to $\frac{9}{8}<\frac{5}{4}$. For $n=3$, we have that $1+\frac{1}{2^3}+\frac{1}{3^3}<\frac{5}{4}-\frac{2}{3^3}$, which is the base case for the induction.

Too easy for JBMO just use integral...