Please check my answer to $\sum\limits_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$

Question

$$\sum_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$$

I have this answer, please let me know if there is a more beautiful proof.

My answer:

at first, we prove two inequalities:

1. If $x\in (x,\pi)$ then $\sin x \leq x$
2. When $x\in(0,\frac{\pi}{2})$, $\sin x \geq \frac{2x}{\pi}$

1) first, let $y = \sin x -x $

$y^{\prime} = \cos x -1 \leq 0$

so $\sin x - x \leq \sin 0 -0 = 0$

which can be rewritten as

$\sin x \leq x$

2) Let $y=\sin x - \frac{2x}{\pi}$

thus $y^{\prime} = \cos x - \frac{2}{\pi}$ because $x\in (0, \frac{\pi}{2})$

so y at first decreases and then increases on the boundary of $x \in (0,\frac{\pi}{2})$

so $ \sin x - \frac{2}{\pi}\leq \max \{{\sin 0 - \frac{2}{\pi}0, \sin (\frac{\pi}{2} - \frac{2}{\pi}\frac{\pi}{2}) \}}$

so $\sin x \leq \frac{2x}{\pi}$

Then select $M\in N$

$\frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m+1} + \ldots + \frac{\sin ((m+n)x)}{m+n} \leq \frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m} + \ldots + \frac{\sin ((m+n)x)}{m} $

=> $\frac{1}{2M} \times \frac{\sin ((m-\frac{1}{2})x) - \sin ((n+\frac{1}{2})x)}{\sin \frac{x}{2}} < \frac{1}{m \times \sin \frac{x}{2}} \times \sin x + \frac{\sin 2x}{2} + \ldots + \frac{\sin ((m-1)x)}{m-1} < x + \frac{2x}{2} + \ldots + \frac{(m-1)x}{m-1} $

so just need to prove that

$(m-1)x + \frac{1}{m \times \sin \frac{x}{2}} \leq 2\sqrt{\pi}$

select M which satisfies

$ \frac{\sqrt{\pi}}{x} \leq m < \frac{\sqrt{\pi}}{x} + 1 $

so $ (m-1)x < [ \frac{\sqrt{\pi}}{x} \times x = \sqrt{\pi} ] $

thus

$\frac {1}{m \times \sin(\frac{x}{2})}\leq[ \frac{1}{\sqrt{\pi}}\times \frac{2}{\frac{ \sin (0.5x)}{0.5x}} = \frac{1}{\sqrt{\pi}} \times \frac{2}{\frac{\sin 0.5x}{0.5x}} ]$

because $x\in (0, \pi)$ thus $\frac{x}{2} \in (0, \frac{\pi}{2})$

$ (m-1)x + \frac{1}{m \times \sin(0.5x)} \leq 2\sqrt{\pi} $

thanks, for viewing and commenting.

ps. I'm still learning latex and mathematics, so my answer isn't pretty to read, nor is the latex I wrote.

Answer 1

I was correct.

Read the comments for better ideas.

Answering this out of a need for my question to have an answer, and I wrote a correct answer in my post.

Yea, for me.