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.
 A: \begin{align*}
\sum_{k=1}^\infty\frac{1}{k^3}&<\sum_{k=2}^\infty\frac{1}{(k-1)(k)(k+1)}+1\\
&=\frac{1}{2}\sum_{k=2}^\infty\left[\frac{1}{k-1}-\frac{2}{k}+\frac{1}{k+1}\right]+1\\
&=\frac{1}{2}\left(1-\frac{1}{2}\right)+1\\
&=\frac{5}{4}
\end{align*}
A: Hint. Note that
$$1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<1+\sum_{n=2}^{2018}\frac{1}{n^3-n}$$
and the sum on the right is easy to be found (it's telescopic).
A: 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.
A: 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 $\zeta(3) < 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{12} = 1.24537 < 5/4$
