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?
As suggested by Greg Martin, the series-integral test is a fast way to go, but you can achieve better estimates with creative telescoping. We have: $$ \zeta(3)=\sum_{n\geq 1}\frac{1}{n^3}=1+\sum_{n\geq 1}\frac{1}{(n+1)^3}\color{red}{\leq} 1+\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)}=1+\frac{1}{4}=\color{red}{\frac{5}{4}}.\tag{1}$$ That can be improved up to: $$ \zeta(3) \color{red}{\leq} 1+\sum_{n\geq 2}\frac{1}{n^3-n/9} = \frac{108\log 3-109}{8}=\color{red}{1.20626}\ldots \tag{2}$$ Another chance is given by the identity (see here, for instance) $$ \zeta(3) = \frac{5}{2}\sum_{n\geq 1}\frac{(-1)^{n-1}}{n^3\binom{2n}{n}}\tag{3}$$ that can be proved through creative telescoping, too.
By just taking the first three terms of the series we get: $$ \zeta(3)\color{red}{\leq} \frac{1039}{864}=\color{red}{1.202546}\ldots\tag{4}$$
We have (for $k\ge 2$) $$\sum_{n=k}^{\infty}\frac1{n^3}<\frac12\sum_{n=k}^{\infty}\frac2{n^3-n}=\frac12\sum_{n=k}^{\infty}\left(\frac1{n(n-1)}-\frac1{n(n+1)}\right)=\frac{1}{2k(k-1)}.$$
$$1+ \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} + \frac{1}{216} + \frac{1}{343} + \frac{1}{512} + \frac{1}{729} + \cdots < 1+ \frac{1}{8} + \frac{1}{8} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{512}+\frac{1}{512}+\cdots$$
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
$$\mathrm{sum} < \color{red}{1.202205}$$
The simplest is a comparison with a lower Riemann sum: $$\sum_{n=2}^\infty\frac1{n^3}\le\int_1^{\infty}\frac{\mathrm d\mkern1mux}{x^3}=\biggl[-\frac1{2x^2}\biggr]_1^{\infty}=\frac12.$$
