# Cannot prove inequality

#### idontknow

Prove that $$\displaystyle \: a_n =\frac{|\sin(n)|}{n} +\frac{|\sin(n+1)|}{n+1}>\frac{1}{6}\; ,n\in \mathbb{N}$$.

#### DarnItJimImAnEngineer

Pretty sure that's false.

1 person

#### romsek

Math Team
Pretty sure that's false.
It's pretty clearly false as $|\sin(x)|\leq 1$ and the denominators can increase without limit.

It looks like it's false $\forall n > 12$

1 person

#### romsek

Math Team
The angels are radians .
The angels are playing their harps.

The angles are in radians.

4 people

#### idontknow

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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#### tahirimanov19

Prove that $$\displaystyle \: a_n =\frac{|\sin(n)|}{n} +\frac{|\sin(n+1)|}{n+1}>\frac{1}{6}\; ,n\in \mathbb{N}$$.
It is true if n<6. Else, it is false.

1 person

#### DarnItJimImAnEngineer

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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#### tahirimanov19

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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