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.

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

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 $$1 + \frac{1}{2^3} + \sum_{2015\geq n\geq 3} \frac{1}{n^3} < \frac{9}{8} + \sum_{n\geq 3} \frac{1}{2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = \frac{9}{8} +\frac{1}{8} = \frac{5}{4}.$$

Split the sum as $1 + 1/8 + \sum_{n=3}^\infty 1/n^3$

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

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

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

Note that for $n>1$,

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

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