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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.

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    $\begingroup$ Did you try upper bounding the sum with integral? $\endgroup$ – Mark May 13 '17 at 17:43
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    $\begingroup$ I just want to say that this problem is meant to be solved by 15 to 16 year olds $\endgroup$ – Luka Markovic May 13 '17 at 17:44
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    $\begingroup$ It should be mentioned that $\lim\limits_{k\to\infty}\sum\limits_{n=1}^k\frac{1}{n^3}=\zeta(3)=1.2020569\dots>\frac{5}{4}$ $\endgroup$ – JMoravitz May 13 '17 at 17:51
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    $\begingroup$ @Mark I see. I'll leave my comment, because it shows why we have to split the sum. $\endgroup$ – Sha Vuklia May 13 '17 at 17:56
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    $\begingroup$ @JMoravitz I believe your inequality is pointing the wrong way. $5/4 = 1.25$ $\endgroup$ – Mark May 13 '17 at 17:59
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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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Split the sum as $1 + 1/8 + \sum_{n=3}^\infty 1/n^3$

You need $\sum_{n=3}^\infty 1/n^3 \lt 1/8$.

$\sum_{n=3}^\infty 1/n^3 \lt \int_{x=2}^\infty 1/x^3 dx = 1/8$

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  • $\begingroup$ Just noticing here that I substituted $2015$ with $\infty$. The problem still follows a fortiori $\endgroup$ – Mark May 13 '17 at 18:06
  • $\begingroup$ This also shows that this contest question can updated and used in perpetuity :-) $\endgroup$ – Mark May 14 '17 at 18:31
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Note that for $n>1$,

$$\frac{1}{n^3}<\frac{1}{n^3-n}=\frac{1}{2(n-1)}-\frac{1}{n}+\frac{1}{2(n+1)}$$

$$\sum_{n=2}^{2015}\frac{1}{n^3-1}=\frac{1}{2(1)}-\frac{1}{2(2)}-\frac{1}{2(2015)}+\frac{1}{2(2016)}=\frac{2031119}{8124480}<\frac{1}{4}$$

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