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.
 A: For every $k \geqslant 2$ and $n \geqslant k$ we have
$$\sum_{m = k}^n \frac{1}{m^3} < \sum_{m = k}^n \int_{m-1}^m \frac{dt}{t^3} = \int_{k-1}^n \frac{dt}{t^3} < \int_{k-1}^\infty \frac{dt}{t^3}.$$
A suitable choice of $k$ yields the desired inequality.
A: 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 $\zeta(3)$.
A: Another way to tackle these problems is by proving a stronger statement using induction.
For example, in this case you might prove that for $n\geq 2$ $$\sum_{j=1}^n\frac{1}{j^3}\leq \frac{5}{4}-\frac{4}{15n^2}.$$
The case $n=2$ is easily verified ($\frac{9}{8}\leq \frac{5}{4}-\frac{1}{21}$). Now for the induction step:
$$\sum_{j=1}^{n+1}\frac{1}{j^3}=\frac{1}{(n+1)^3}+\sum_{j=1}^{n}\frac{1}{j^3}\leq \frac{1}{(n+1)^3}+\frac{5}{4}-\frac{4}{15n^2}$$ and now what we have to verify is that $$\frac{1}{(n+1)^3}+\frac{5}{4}-\frac{4}{15n^2}\leq \frac{5}{4}-\frac{4}{15(n+1)^2}.$$
This is equivalent to $$\frac{1}{(n+1)^3}\leq \frac{4}{15}\cdot\frac{2n+1}{n^2(n+1)^2}$$ $$\frac{4}{15}\geq\frac{n^2}{(n+1)(2n+1)}$$ which is true because we have equality in $n=2$ and the function on the RHS is decreasing.
Actually, the most natural way to proceed is to write $S_n\leq \frac{5}{4}-\frac{A}{n^s}$ and to go the above steps to find $A$.
