Let $n$ be a natural number and let $0\lt x\lt{\pi}$. Then, here are my questions.
Question 1: Is the following true?
$$\sum_{k=1}^{n}\frac{\cos(kx)}{k}\gt -1$$
Question 2: Is the following true?
$$\sum_{k=1}^{n}\frac{\sin(kx)}{k}\gt0$$
This is a possible hint for solution; perhaps someone can finish it along these lines (it won‘t fit as a comment). We have
$$\sin x+\dfrac{\sin 2x}{2}+\dfrac{\sin 3x}{3}+\ldots+ \dfrac{\sin nx}{n}=\sum_{k=1}^n\int_0^x\cos kt\,dt,$$
$$2\sum_{k=1}^n\cos kt=\sin((n+1/2)t)/\sin(t/2)-1$$ (by taking the real part of $\sum_{k=1}^n e^{ikt}$)
so we want to show
$$\int_0^x(\frac{\sin(n+1/2)t}{\sin(t/2)}-1)dt>0.$$
It‘s easy without that $-1$ (as $1/\sin (t/2)$ decreases); to do it really (with $-1$) a better estimate is needed.
TEOREMA (L.Fejer-1910, D. Jackson-1912, T.J.Gronwall- 1912). Pentru $x\in (0,\pi)$ au loc inegalitatile
\[(1)\; \; \; \; \; \; \begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin{kx}}{k}> 0 \\ \\ \hline \end{array}\; \; ,\; \; \forall n \in{\mathbb N}. \]
DEMONSTRATIA I .
Prin efectuarea unor calcule elementare se constata
\[\frac{d}{d\varphi}\left\{ \frac{\sin{2k\phi}}{(\sin{\phi})^{2k}}\right\}=-2k\frac{\sin{(2k-1)\phi}}{(\sin{\phi})^{2k+1}}\; . \]
Efectuand calculele precum si substitutia $\phi \leadsto \frac{x}{2}\; ,\;$ prin integrare avem
\[\frac{\sin{kx}}{k}=2\left(\sin{\frac{x}{2}}\right)^{2k}\int_{\frac{x}{2}}^{\frac{\pi}{2}}\frac{\sin{(2k-1)\phi}}{(\sin \phi)^{2k+1}}\; d \phi \; \; ,\; k\in{\mathbb N}. \]
Insumand pentru $k\in \{1,2,...,n\}$ se obtine
\[\sum_{k=1}^{n}\frac{\sin{kx}}{k}=2\int_{\frac{x}{2}}^{\frac{\pi}{2}}\sum_{k=1}^{n}\left[r(x,\phi)\right]^{k}\frac{\sin{(2k-1)\phi }}{\sin{\phi}}\; d\phi \]
unde \[r(x,\phi): =\left(\frac{ \sin\frac{x}{2}}{\sin{\phi}}\right)^{2}\in [0,1] \]
pentru $0\le \frac{x}{2}\le \phi<\frac{\pi}{2}\; .$ Egalitatea de mai sus implica
\[\sum_{k=1}^{n}\frac{\sin{kx}}{k}=\int_{x}^{\pi}\sum_{k=1}^{n}\left[r(x,\frac{\psi}{2})\right]^{k}\frac{\sin{\left(k-\frac{1}{2}\right)\psi }}{\sin{\frac{\psi}{2}}}\; d\psi\; . \]
Aplicand identitatea lui Abel ("‘insumarea prin parti"), adica
\[\sum_{k=1}^{n}A_{k}B_{k}=\sum_{k=1}^{n-1}\left(A_{k}-A_{k+1}\right)\sum_{j=1}^{k}B_{j}+A_{n}\sum_{j=1}^{n}B_{j}\; , \]
si notand
\[\left\{\begin{array}{rcl}{\mathbf r}&: =&r(x,\frac{\psi}{2})\; \; , \; \;{\mathbf r}\in [0,1]\; \; \; \mbox{vezi}\; \; ()\\{\mathcal F}_{n}(\psi)&: =&\sum_{k=1}^{n}\sin{\left(k-\frac{1}{2}\right)\psi}\; \; \; , \; \;{\mathcal F}_{n}(\psi)\ge 0 \; \; \mbox{dac‘a}\; x\in [0,\pi]\; \;-\mbox{vezi L. Fej\‘er}\end{array}\right.\; , \] din identitatea lui Abel g‘asim
\[\sum_{k=1}^{n}\frac{\sin{kx}}{k}=\int_{x}^{\pi}\left\{(1-{\mathbf r})\sum_{k=1}^{n-1}{\mathbf r}^{k-1}{\mathcal F}_{k}(\psi)+{\mathbf r}^{n}{\mathcal F}_{n}(\psi)\right\}\frac{d\psi}{\sin{\frac{\psi}{2}}}\; . \]
Deoarece ${\mathcal F}_{n}(\psi)=\frac{1-\cos(n\psi)}{2\sin{\frac{\psi}{2}}}\; ,$ concludem cu
\[\begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin{kx}}{k}=\frac{1}{2}\int_{x}^{\pi}\left\{(1-{\mathbf r})\sum_{k=1}^{n-1}{\mathbf r}^{k-1}\left(1-\cos(k\psi)\right)+{\mathbf r}^{n}(1-\cos(n\psi))\right\}\frac{d\psi}{\left(\sin{\frac{\psi}{2}}\right)^{2}}\\ \\ \hline \end{array}\; . \]
Aceasta reprezentare completeaza demonstra‘tia I.
DEMONSTRATIA II.
Fie $P_{n}(x)$ polinomul lui Legendre de gradul $n$, adic‘a
\[\begin{array}{c}P_{n}(x)=\frac{1}{n! 2^{n}}\left[(x^{2}-1)^{n}\right)^{(n)}={}_{2}F_{1}\left(-n,n+1;1;\frac{1-x}{2}\right)=\\ \\ =\frac{1}{2^{n}}\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}{n\choose 2k}(-1)^{k}\frac{(n+1)_{n-2k}(2k)!}{k!(n-k)!}x^{n-2k}\end{array}\]
unde
\[\begin{array}{c}{}_{2}F_{1}(-n;b;c;z): =\sum_{k=0}^{n}(-1)^{k}{n\choose k}\frac{(b)_{k}}{(c)_{k}}z^{k}\\ (d)_{k}: =d(d+1)\cdots (d+k-1)\; ,\; \; k\in{\mathbb N}\; ,\; (d)_{0}: =1 . \end{array}\]
Se cunosc urmatoarele:
-- radacinile lui $P_{n}(x)$ sunt reale,distincte, situate in $(-1,1)$;
-- $|P_{n}(t)|\le 1 \; ,\; \; \forall t\in [-1,1] .$
Demonstratia a II-a (vezi [16] precum si comentariile lui R.Askey )
se bazeaza pe identitatea:
\[\begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin(k\cdot\arccos{x})}{k}= \frac{\sqrt{1-x}}{2}\int_{-1}^{x}\frac{1-P_{n}(y)}{1-y}\frac{dy}{\sqrt{x-y}}\\ \\ \hline \end{array}\; ,\; x\in (-1,1). \; .\]
Observatii.
1) Inegalitatea (1) a fost conjecturata de catre Leopold (Lipot) Fejer. Ulterior a fost demonstrata de catre D.Jackson-[14] si T.H.Gronwall-[12].
In prezent se cunosc peste 100 de demonstratii. Am ales pentru a Va prezenta pe cele pe care subsemnatul le considera "mai simple".
Se spune ca pana la moasrte L.Fejer a cautat sa gaseasca noi demonstratii a lui (1). L.Fejer a predat si la Universitatea din Cluj (sub numele de L.Weiss).
2)Desi simpla la prima vedere, inegalitatea (1) a dat de furca matematicienilor.
3) Inegalitatea (1) intervine in urmatoarele domenii: Serii Fourier (fenomenul lui Gibbs - vezi [13]), Polinoame ortogonale ([1]-[5],[18] ), Functii Complexe (Demonstratia conjecturii lui Bieberbach,functii univalente, [4],[6]), Teoria Aproximarii ([5]).
BIBLIOGRAFIE.
[1]R. Askey , Orthogonal Polynomials and Special Functions,
Regional Conf.Lect.Appl.Math., vol.21,SIAM,Philadelphia,Pa., 1975.
[2]R.Askey Positive quadrature methods and positive polynomial sums, ‘in Approximation Theory V, Academic Press, 1986.
[3] R. Askey and J. Fitch , Integral reprezentations for Jacobi
polynomial amd some applications ,J.Math.Anal.Appl., 26 (1969)
411-437.
[4]R. Askey and G. Gasper , Positive Jacobi polynomial sums,(II),
Amer.J.Math., 98 (1976) 709-737.
[4]R. Askey and G. Gasper , Inequalities for polynomials}, ‘in "The Bieberbach Conjecture", Proc.of the Symposium on the Occasion
of the Proof, Mathematical Surveys and Monographs, 21, Amer.Mathematical Society, 1986, 7-32.
[5]H. Bavinck , Jacobi Series and Approximation,
Mathematical Centre Tracts 39, Mathematisch Centrum Amsterdam 1972.
[6] L. de Branges , The Story of the Verification of the Bieberbach Conjecture, ‘in " The Bieberbach Conjecture", Proc.of the Symposium on the Occasion of the Proof, Mathematical Surveys and Monographs, 21, Amer.Mathematical Society, 1986, 199-203.
[7] L. Fejer , Sur les fonctions bornees et integrables,
C.R.Acad.Sci.Paris 131 (1900) 984-987.
[8] L. Fejer , Sur le develpopment d‘une fonction arbitraire suivant les fonctions de Laplace, C.R.Acad.Sci.Paris , 146 (1908) 224-227.
[9]L. Fejer , Ueber die Laplacesche Reihe, Math. Ann. (1909)76-109.
[10] L. Fejer , Einige Saetze , die sich auf das Vorzeichen einer ganzen rationalen Funktion beziehen u.s.w., Monta.fuer Math. und Phys., 35 (1928) 305-344.
[11]L. Fejer , Gesammelte Arbeiten (I)-(II), Birkhauser Verlag, Basel, 1970 .
[12]T.H. Gronwall , Ueber die Gibbsche Erscheinung und die trigonometrischen Summen $\sin{x}+(1/2)\sin{2x}+...+(1/n)\sin{nx}$ , Math.Ann., 72 (1912) 228-243.
[13] E. Hewitt and R.E. Hewitt , The Gibbs-Wilbraham phenomenon: an epsiode in Fourier analysis, Arch.Hist.Exact Sci., 21 (1979) 129-160.
[14]D. Jackson , Ueber eine trigonometrische Summe, Rend.Circ.Mat.Palermo 32(1911) 257-262.
[15]A. Lupas , Advanced Problem 6517, Amer.Math.Monthly (1986) p. 305 ; (1988) p.264.
[16]A. Lupas , Advanced Problem 6585, Amer.Math.Monthly
(1988) p. 880 ; (1990) p.859-860.
[17]L. Lupas , An identity for ultraspherical polynomials, Revue d${}^{,}$Analyse numerique et de Theorie de l‘approximation, tome 24 , no.1-2 (1995) 181-185.
[18]G. Szego , Orthogonal Polynomials , Amer.Math.Soc.Colloq. Publications vol.23, fourth ed.,Amer.Math.Soc., Providence, R.I., 1975.
C?lcat-up inegalitate $ \sum_{k=1}^{n}\frac{\sin{kx}}{k}>x\left(1-\frac{x}{\pi}\right)^3$ de?ine pentru $ x \in (0,\pi).$
参考这里Sawtooth wave
**In short:** let $f_n(x)$ denote the function on the lhs of the inequality. Of course, $f_1(x)=\sin x\geq 0$ on $[0,\pi]$. We will prove that $f_n(x)\geq 0$ on $[0,\pi]$ by induction on $n$. It is not too hard to determine the local minima of $f_n$ on $[0,\pi]$ by investigating its derivative. Then Ma Ming observed that $f_n$ coincides with $f_{n-1}$ on these local minima. And the induction step follows easily. Of course, $f_n(0)=f_n(\pi)=0$. We will actually prove that
>$$
f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k}>0\qquad\forall x\in(0,\pi).
$$
**Remark:** it is worth noting that the $f_n$‘s are the partial sums of the Fourier series of the same sawtooth function. Just [look at the case $n=6$][1], for instance, to see how they tend to approximate it nicely. [See here][2] to get an idea how to estimate the error in such approximations. As pointed out by math110, there are many proofs of this so-called Fejer-Jackson inequality. It can even be shown that the [$f_n$‘s are bounded below][3] by a certain nonnegative polynomial on $[0,\pi]$. The proof below is at the calculus I level. I‘m not sure it can be made more elementary.
**Proof:** first, $f_1(x)=\sin x$ is positive on $(0,\pi)$. Assume this holds for $f_{n-1}$ for some $n\geq 2$. Then observe that $f_n$ is differenbtiable on $\mathbb{R}$ with
$$
f_n‘(x)=\sum_{k=1}^n\cos kx=\mbox{Re} \sum_{k=1}^n (e^{ix})^k.
$$
For $x\in 2\pi \mathbb{Z}$, we have $f_n‘(x)=n$. So the zeros of $f_n‘$ are the zeros of
$$
\mbox{Re}\;e^{ix}\frac{e^{inx}-1}{e^{ix}-1}=\mbox{Re}\;e^{i(n+1)x/2}\frac{\sin (nx/2)}{\sin(x/2)}=\frac{\cos((n+1)x/2)\sin (nx/2)}{\sin(x/2)}.
$$
This yields
$$
\frac{nx}{2}\in \pi\mathbb{Z}\quad\mbox{or}\quad \frac{(n+1)x}{2}\in \frac{\pi}{2}+\pi\mathbb{Z}
$$
i.e.
$$
x\in \frac{2\pi}{n}\mathbb{Z}\quad\mbox{or}\quad x\in \frac{\pi}{n+1}+\frac{2\pi}{n+1}\mathbb{Z}.
$$
Between $0$ and $\pi$, these are ordered as follows:
$$
0<\frac{\pi}{n+1}<\frac{2\pi}{n}<\frac{3\pi}{n+1}<\frac{4\pi}{n}<\ldots < \frac{2\lfloor n/2\rfloor \pi}{n}\leq \pi.
$$
The sign of $f_n‘$ changes at each of these zeros, starting from a positive sign on $(0,\pi/(n+1))$. It follows that $f_n$ is positive on the latter, positive on the last interval (if nontrivial, i.e. in the odd case), with local minima at
$$\frac{2j\pi}{n}\qquad\mbox{for}\qquad j=1,\ldots,\lfloor n/2\rfloor.$$
But now here is Ma Ming‘s key observation: for these values, we have
$$
f_n\left(\frac{2j\pi}{n}\right)=f_{n-1}\left(\frac{2j\pi}{n}\right)+\sin\left(n\cdot\frac{2j\pi}{n}\right)=f_{n-1}\left(\frac{2j\pi}{n}\right)>0
$$
by induction step. It follows that $f_n(x)>0$ on $(0,\pi)$. QED.
[1]: http://www.wolframalpha.com/input/?i=sin%20%28x%29%2bsin%20%282x%29/2%2bsin%20%283x%29/3%2bsin%20%284x%29/4%2bsin%285x%29/5%2bsin%286x%29/6
[2]: https://math.stackexchange.com/questions/57054/asymptotic-error-of-fourier-series-partial-sum-of-sawtooth-function
[3]: https://math.stackexchange.com/questions/177995/a-pseudo-fejer-jackson-inequality-problem
How to prove that $\forall x \in \mathbb{R}$, $n \in \mathbb{N}$, we have
\begin{align}
\sum_{k=1}^{n}\frac{|\sin{kx}|}{k}\ge |\sin{nx}| ?
\end{align}
Given $n\ge 1$ and $x\in \Bbb R$, denote
$$f_n(x)=\sum_{k=1}^n\frac{|\sin kx|}{k} \quad\text{and}\quad g_n(x)=f_n(x)-|\sin nx|.$$
----------
> **Lemma:** For every $n\ge 1$,
>
> (i) $f_n$ is increasing on $[0,\frac{\pi}{n+1}]$;
> (ii) $g_n$ is increasing on $[0,\frac{\pi}n]$;
> (iii) $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.
**Proof of Lemma:** Note that when $x\in [0,\frac{\pi}n]$,
$$f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k},\quad\text{and}\quad g_n(x)=f_n(x)-\sin nx,$$
so
$$f_n‘(x)=\sum_{k=1}^n \cos kx \quad\text{and}\quad g_n‘(x)=f_n‘(x)-n\cos nx.$$
(i) Given $x\in[0,\frac{\pi}{n+1}]$, noting that $\cos\frac{kx}{2}\ge 0$ for $k=0,\pm1,\dots, \pm (n+1)$, we have
$$f_n‘(x)=\sum_{k=1}^n \frac{\cos kx +\cos(n+1-k)x}{2}=\cos \frac{(n+1)x}{2} \cdot\sum_{k=1}^n \cos\frac{(n+1-2k)x}{2}\ge 0.$$
(ii) Since the cosine function is decreasing on $[0,\pi]$, when $x\in [0,\frac{\pi}{n}]$, $\cos k x\ge \cos nx$ for $k=1,\dots,n$, so $g‘(x)\ge 0$.
(iii) When $n=1$, the statement is clearly true; when $n=2$, since $f_2$ is concave on $[\frac{\pi}{4},\frac{\pi}{2}]$, $f_2(\frac{\pi}{4})>1$ and $f_2(\frac{\pi}{2})=1$, the statement is also true. By induction, we may assume that $f_{n-1}\ge 1$ on $[\frac{\pi}{2(n-1)},\frac{\pi}{2}]$ for some $n \ge 3$, and the conclusion $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$ follows from the facts below. Firstly, $f_n\ge f_{n-1}$; secondly, $f_n$ is increasing on $[0,\frac{\pi}{n+1}]\supset [\frac{\pi}{2n},\frac{\pi}{2(n-1)}]$; thirdly,
$$\sin \frac{\pi t}{2}\ge t,\ \forall t\in[0,1]\Longrightarrow f_n(\frac{\pi}{2n})=\sum_{k=1}^n\frac{\sin \frac{k\pi}{2n}}{k}\ge 1.\qquad \square$$
----------
Now we can prove that $g_n\ge 0$ by using the lemma. Since $g_n(\pi\pm x)=g_n(x)$, we may focus on $x\in[0,\frac{\pi}{2}]$. Since $g_n(0)=0$, by (ii), we know that $g_n(x)\ge 0$ on $[0,\frac{\pi}{n}]$. Since $g_n\ge f_n -1$, by (iii) we know that $g_n\ge 0$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.
参考：这里
原文地址：https://www.cnblogs.com/Eufisky/p/8995608.html