(Fejer-Jackson-Growall不等式) 1910年Fejer猜想,三角函数级数

\[\frac{\pi-x}{2}=\sum_{k=1}^{\infty}\frac{\sin kx}{k},\quad 0<x\leq\pi\]

的所有部分和

\[S_n(x)=\sum_{k=1}^{n}\frac{\sin kx}{k}>0,\quad n=1,2,\ldots,0<x<\pi\]

(Turan, 1952)设$\{a_k\}(k=1,2,\ldots)$为正的严格递减数列.设$\displaystyle S_m=\sum_{k=1}^{m}b_k\geq 0$.若有$S_m>0$,则有\[\sum_{k=1}^{n}a_kb_k=\sum_{k=1}^{n-1}S_k(a_k-a_{k+1})+a_nS_n>0.\]

利用

\begin{align*}

\sum_{k=1}^n{\sin \left( 2k-1 \right) x}&=\sum_{k=1}^n{\frac{\cos 2\left( k-1 \right) x-\cos 2kx}{\text{2}\sin x}}

\\

&=\frac{1-\cos 2nx}{\text{2}\sin x}=\frac{\sin ^2nx}{\sin x}

\\

\frac{d}{dt}\left[ \frac{\sin 2kt}{2k\left( \sin t \right) ^{2k}} \right] &=-\frac{\sin \left( 2k-1 \right) t}{\left( \sin t \right) ^{2k+1}},

\end{align*}

可知

\[\frac{\sin kx}{k}=2\int_{x/2}^{\pi /2}{\left( \frac{\sin \frac{x}{2}}{\sin \theta} \right) ^{2k}\frac{\sin \left( 2k-1 \right) \theta}{\sin \theta}d\theta},\]

因此

\[\sum_{k=1}^n{\frac{\sin kx}{k}}=2\int_{x/2}^{\pi /2}{\left[ \sum_{k=1}^n{\left( \frac{\sin \frac{x}{2}}{\sin \theta} \right) ^{2k}\frac{\sin \left( 2k-1 \right) \theta}{\sin \theta}} \right] d\theta}.\]

记

\[r^{2k}=\left( \frac{\sin \frac{x}{2}}{\sin \theta} \right) ^{2k},\quad k=1,2,\ldots\]

明显$r^{2k}$为递减的,当$0<x<\pi,\frac{x}{2}\leq\theta\leq\frac{\pi}{2}$时,有

\[\sum_{k=1}^n{r^{2k}\sin \left( 2k-1 \right) \theta}>0,\qquad 0<\theta<\pi,\]

由此得证.