Cannot prove inequality

Dec 2015
975
128
Earth
Prove that \(\displaystyle \: a_n =\frac{|\sin(n)|}{n} +\frac{|\sin(n+1)|}{n+1}>\frac{1}{6}\; ,n\in \mathbb{N}\).
 
Dec 2015
975
128
Earth
The original inequality is \(\displaystyle a_n =\sum_{i=n}^{2n} \frac{|\sin(i)|}{i}>\frac{1}{6}\).n-radian
Prove that \(\displaystyle a_n = \frac{|\sin(n)|}{n} + \frac{|\sin(n+1)|}{n+1} +...+\frac{|\sin(2n)|}{2n}>\frac{1}{6} \; , n\in \mathbb{N}\). n-radian
But I'm not being able to prove it. Any hint?
 
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Jun 2019
493
262
USA
Prove that \(\displaystyle a_n=\displaystyle \frac{|\sin(n)|}{n} +\frac{|\sin(n+1)|}{n+1} +...+\frac{|\sin(2n)|}{2n}>\frac{1}{6} \; , n\in \mathbb{N}\). n-radian
But I'm not being able to prove it. Any hint?
I would say compare it to $\displaystyle \int_n^{2n} \frac{|\sin~x|}{x} dx$, but the function is not monotonically decreasing, so I don't know how to prove the average value of |sin i| is large enough.

I can tell you practically the sums seem to converge to about 0.4413 for large n, but I don't know how to help with a rigorous proof.
 
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Mar 2015
182
68
Universe 2.71828i3.14159
The original inequality is \(\displaystyle a_n =\sum_{i=n}^{2n} \frac{|\sin(i)|}{i}>\frac{1}{6}\).n-radian
Prove that \(\displaystyle a_n = \frac{|\sin(n)|}{n} + \frac{|\sin(n+1)|}{n+1} +...+\frac{|\sin(2n)|}{2n}>\frac{1}{6} \; , n\in \mathbb{N}\). n-radian
But I'm not being able to prove it. Any hint?
$$\left( \prod_{k=n}^{2n} \dfrac{|\sin(k)|}{k} \right)^{\frac{1}{n+1}} \approx 1?$$
 
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