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?
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.
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?