TITLE: Prove $\left|\sum\limits_{k=2001}^{m}a_{k}\sin{(kx)}\right|\le 1+\pi $ ,$m\ge 2001,x\in R$
QUESTION [13 upvotes]: let $\{a_{n}\}$ is non-increasing postive sequence;show that
if for $n\ge 2001,na_{n}\le 1$, then for any positive integer numbers $m\ge 2001,x\in R$, we have
$$\left|\sum\limits_{k=2001}^{m}a_{k}\sin{(kx)}\right|\le 1+\pi $$
This problem is 2001 china team problem.http://www.doc88.com/p-30356357991.html
  It is well known that
Fejér-Jackson-Gronwall inequality:
$$\sum\limits_{k=1}^{n}\dfrac{\sin{kx}}{k}>0$$
Inequality $\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0$ (Fejer-Jackson)
and we know that have
$$\left|\sum\limits_{k=1}^{n}\dfrac{\sin{(kx)}}{k}\right| \le 2\sqrt{\pi}$$
Please check my answer to $\sum\limits_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$
and the simple link:
How prove this inequlity $\sum_{k=n}^{2n-3}\frac{|\sin{k}|}{k}<\frac{1}{\sqrt{2}},n\ge 3$
Proving that $\sum_{k=n}^{2n-2} \frac{|\sin k\ |}{k} < 0.7\ln 2$ for $n\ge2$
But for my question How prove it?  Thank you

REPLY [4 votes]: If we first rearrange the sum in the following way:
$$ \sum_{k=2001}^m a_k\sin(k x) = a_m \sum_{k=2001}^{m}\sin(k x)+(a_{m-1}-a_m)\sum_{k=2001}^{m-1}\sin(k x)+\ldots+(a_{2002}-a_{2003})\sum_{k=2001}^{2002}\sin(k x)+(a_{2001}-a_{2002})\sin(2001x),$$
then compute:
$$ \sum_{k=2001}^m \sin(kx) = \frac{\cos\left(\frac{4001x}{2}\right)-\cos\left(\frac{(2m+1)x}{2}\right)}{2\sin(x/2)},$$
we have:
$$ \sum_{k=2001}^m a_k\sin(k x) = a_{2001}\cdot\frac{\cos\left(\frac{4001x}{2}\right)-1}{2\sin(x/2)}+a_m\cdot \frac{1-\cos\left(\frac{(2m+1)x}{2}\right)}{2\sin(x/2)}+\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{1-\cos\left(\frac{(2k+1)x}{2}\right)}{2\sin(x/2)}.$$
Now, since the function $\frac{1-\cos\left(\frac{Nx}{2}\right)}{2\sin(x/2)}$ 
is non-negative and bounded by $\frac{N}{I_0(1)}<\frac{4N}{5}$, we have:
$$ \sum_{k=2001}^m a_k\sin(k x) \leq 2m a_m+\sum_{k=2001}^{m-1}2k(a_k-a_{k+1})\leq 2+2\left(\sum_{k=2001}^{m-1}k a_k-\sum_{k=2001}^{m-1}k a_{k+1}\right),$$
from which:
$$\left|\sum_{k=2001}^m a_k\sin(k x)\right| \leq 2+2\log\left(\frac{m}{2001}\right)$$
follows. This is clearly not the theorem's claim, since it depends on $m$. 
Anyway I hope that my argument can be refined in order to remove that dependence.
A refined inequality is:
$$ 0\leq \frac{1-\cos\left(\frac{2n+1}{2}x\right)}{2\sin(\frac{1}{2}x)}\leq\min\left(\frac{(2n+1)^2}{8}x,\frac{1}{\sin(x/2)}\right).$$
By symmetry, we only need to prove the inequality for $x\in[0,\pi]$.
If $x=\frac{4}{2h+1}$ with $h\geq m$, then:
$$\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{1-\cos\left(\frac{(2k+1)x}{2}\right)}{2\sin(x/2)}\leq \frac{4}{2h+1}\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{(2k+1)^2}{8}\leq \frac{2(m-k)}{2h+1}\leq 1.$$
If $x=\frac{4}{2h+1}$ with $h\in(2001,m-1]$, then:
$$\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{1-\cos\left(\frac{(2k+1)x}{2}\right)}{2\sin(x/2)}\leq 1+\frac{a_h-a_m}{2\sin(x/2)}\leq 1+\frac{1}{2h\sin\left(\frac{2}{2h+1}\right)}\leq 2,$$
so the inequality is true in the range $x\in\left[0,\frac{4}{2m+1}\right]$ and we can assume that $x$ is big enough, i.e. $x\geq\frac{4}{4003}$. This gives:
$$\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{1-\cos\left(\frac{(2k+1)x}{2}\right)}{2\sin(x/2)}\leq\frac{a_{2001}}{2\sin\left(\frac{2}{2003}\right)}\leq\frac{1}{2002 \sin\left(\frac{2}{2003}\right)}\leq 1.$$
Putting all together, we have:
$$ \left|\sum_{k=2001}^m a_k \sin(kx)\right|<4+\frac{1}{2001}<1+\pi.$$