<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] 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).
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