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.

By creative telescoping / Euler's acceleration method
$$ \zeta(3)=\sum_{n\geq 1}\frac{1}{n^3} = \frac{5}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3 \binom{2n}{n}}\tag{1} $$
and since the last series is rapidly convergent and with alternating signs,
$$ \zeta(3) \leq \frac{5}{2}\sum_{n=1}^{3}\frac{(-1)^{n+1}}{n^3 \binom{2n}{n}} = \frac{1039}{864}=[1;4,1,14,1,10]\leq [1;4]=\frac{5}{4}.\tag{2}$$
$(1)$ has been historically crucial in Apery's proof of $\zeta(3)\not\in\mathbb{Q}$, for instance.
