How to prove that $\sum_{1}^{\infty} \frac{1}{n^3} \le 1.5$ I have this sequence:
 $$\sum_{1}^{\infty} \frac{1}{n^3}$$
and I need to prove that:  $\sum_{1}^{\infty} \frac{1}{n^3} \le 1.5$
So basically I know that this sequence converges using the integral test, but I don't know how to prove the above statement.
Some help?

$$1+ \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} + \frac{1}{216} + \frac{1}{343} + \frac{1}{512} + \frac{1}{729} + \cdots < 1+ \frac{1}{8} + \frac{1}{8} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{512}+\frac{1}{512}+\cdots$$
In fact, this gets you $$\mathrm{sum} < \color{red}{1 \frac{1}{3} \approx 1.333}$$

And one could easily improve it by changing the first $\frac{1}{8}$ to  $\frac{1}{27}$, this is the largest overestimation. This actually gives:
$$\mathrm{sum} < \color{red}{1 \frac{53}{216} \approx 1.245}$$
By summing the first 31 terms and then using above technique, we can reach 
$$\mathrm{sum} < \color{red}{1.202205}$$
