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

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

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

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.

-
You might include a reference to where you've seen it done with algebra, so we might know what you don't want. –  Gerry Myerson Apr 11 '12 at 0:08
What's the meaning of "with algebra", and of "without algebra"? –  Martin Argerami Apr 11 '12 at 1:56
@MartinArgerami When I was shown the answer, the guy just use basic algebra manipulation to show the result. –  yiyi Apr 11 '12 at 2:02
@Mao Maybe you can use $$\int\limits_0^\infty \frac {\sin (xt)}{t} dt =\frac{\pi}{2};\forall x$$ –  Pedro Tamaroff Apr 11 '12 at 2:45
Probably relevant: $\sum_{i=1}^n \frac{\sin(ix)}{i}$ is the $n$'th partial sum of the Fourier series for $f(x) = \pi/2 - x/2$ on $[0,2\pi]$. The fact that the maximum of this function is somewhat larger than $\pi/2$ is related to the Gibbs phenomenon. I'm not sure where $2 \sqrt{\pi}$ would come in though, it's certainly not a tight bound. –  Robert Israel Apr 11 '12 at 7:46

I was correct.

The best it is possible to state is: $$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots$$ Call $f_N(x)=\sum_{n=1}^N \frac{\sin(nx)}{n}$: it is a $2\pi$-periodic function converging to $\frac{\pi-x}{2}$ in $L_2\left([0,2\pi]\right)$. Since: $$\frac{d f_N(x)}{dx}= \frac{\cos\left(\frac{N+1}{2}x\right)\sin\left(\frac{N}{2}x\right)}{\sin\left(\frac{x}{2}\right)},$$ we know that $f_N(x)$ has $2n$ stationary points in $[0,2\pi]$, local maxima in $x=\frac{(2k+1)\pi}{N+1}$, the first one occurring in $x_N=\frac{\pi}{N+1}$. Once we prove that the value of $f_N(x)$ in any other local maximum is less than $f_N(x_N)$, and that $f_N(x_N)$ is an increasing sequence (I still must find a convincing proof of this two facts, but they look not too hard to deal with and strongly supported by computer inspection) the best bound we can hope in is: $$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\lim_{N\to +\infty}\sum_{n=1}^{N}\frac{\sin\left(\frac{\pi x}{N+1}\right)}{n},$$ where the RHS a Riemann sum associated with: $$\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots<\frac{13}{7},$$ QED.