<s>[INST] Algebra: Prove inequality $\sum_{n=1}^{2015} \frac1{n^3} < \frac 54$ Can someone prove inequality (n is natural):$$\sum_{n=1}^{2015} \frac{1}{n^3} < \frac 5 4$$
I have tried some predictions like $a^3 > a(a - 1)(a - 2) $ but couldn't get anything out of them.
 [/INST] You can use (for $n \geq 2$)
$$\frac{1}{n^3} < \frac{1}{n^2(n-1)} < \frac12\frac{2(n-1) +1}{n^2 (n-1)^2} =\frac{1}{2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2}\right).$$
Write your sum as
$$1 + \frac{1}{2^3} + \sum_{2015\geq n\geq 3} \frac{1}{n^3} < \frac{9}{8} + \sum_{n\geq 3} \frac{1}{2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2}\right)
= \frac{9}{8} +\frac{1}{8} = \frac{5}{4}. $$
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