Proving a progression inequality Prove that $$1+ \frac{1}{2^3} + \cdot \cdot \cdot + \frac{1}{n^3} < \frac{5}{4}  $$
I got no idea of how to approach this problem.

Since:
$$\begin{eqnarray*} \frac{1}{n^3} &=& \frac{1}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)+\frac{1}{2}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\frac{2n+1}{2n^3(n+1)^3}\\&\leq&\frac{1}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)+\frac{1}{2}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\frac{1}{4}\left(\frac{1}{n^4}-\frac{1}{(n+1)^4}\right)\tag{1}\end{eqnarray*}$$

$$ \zeta(3) \leq \frac{1}{2}+\frac{1}{2}+\frac{1}{4}=\frac{5}{4} $$

follows by creative telescoping. With an extra step we may also prove $\frac{6}{5}\leq\zeta(3)\leq\frac{5}{4}$.
More tight approximations (both upper and lower bounds) can be derived from the identity:
$$ \zeta(3) = \sum_{n\geq 1}\frac{5(-1)^{n+1}}{2n^3\binom{2n}{n}}\tag{2}$$
that was crucial in Apery's proof of the irrationality of $\zeta(3)$.
