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.

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 $A$.
