<s>[INST] Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$ Question: Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$
Attempt: When I tried solving this problem, I paired up $1$ with $\frac{1}{2018^3}$, $\frac{1}{2^3}$ with $\frac{1}{2017^3}$ etc. but it wasn't useful.
 [/INST] Another method that is generally applicable:
$\zeta(2n+1) = 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \frac{1}{4^{2n+1}} + \ldots$
$<  1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \left(\frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}}\right) + \ldots $
$= 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \frac{1}{4^{2n}} + \frac{1}{8^{2n}} + \frac{1}{16^{2n}} + \ldots $
$ = 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \frac{1}{2^{2n}(2^{2n} -1)}$
For $n= 1$, we get $\zeta(3) < 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{12} = 1.24537 < 5/4$
</s>