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Here is an exercise, on analysis which i am stuck.

  • How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on $\mathbb{R}$?
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First, let's note that for $x\in(0,2\pi)$ $$\sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\int_{0}^{x}\sum\limits_{k=1}^{n}\cos kt\ dt= -\frac{x}{2}+\int_{0}^{x}\frac{\sin \frac{(2n+1)t}{2}}{2\sin{\frac{t}{2}}}\ dt$$ $$=-\frac{x}{2}+\int_{0}^{x}\left(\frac{1}{2\sin{\frac{t}{2}}}-\frac{1}{t}\right)\sin \frac{(2n+1)t}{2}\ dt +\int_{0}^{x}\frac{\sin \frac{(2n+1)t}{2}}{t}dt.$$ Now, the first integral at the right-hand side tends to zero by the Riemann-Lebesgue lemma. The second one is equal to (via a substitution $s=(2n+1)t/2$) the integral $$\int_{0}^{\frac{(2n+1)x}{2}}\frac{\sin s}{s}\ ds\to\int_{0}^{\infty}\frac{\sin s}{s}\ ds=\frac{\pi}{2}.$$ Therefore $$\lim\limits_{n\to\infty}\ \sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2}=f(x),\qquad x\in(0,2\pi).$$ The series converges on $\mathbb R$ to the periodic extension of $f(x)$.

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  • $\begingroup$ if you take $x=0$ then the left side is a summation of zeros and the right isn't zero. $\endgroup$ – Ofir Dec 8 '10 at 6:51
  • $\begingroup$ The argument works for $x\neq 2\pi m$. $\endgroup$ – Andrey Rekalo Dec 8 '10 at 7:09
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    $\begingroup$ @Prometheus: At that point the series converges to $(\lim_{x\to0,x>0}f(x)+\lim_{x\to0,x<0}f(x))/2=$ $(\lim_{x\to0,x>0}f(x)+\lim_{x\to2\pi,x<2\pi}f(x))/2 =$ $(\frac{\pi}{2}+\frac{\pi-2\pi}{2})/2=0$. $\endgroup$ – AD. Dec 8 '10 at 11:41
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    $\begingroup$ @Prometheus Read about Dirichlet's theorems on Fourier series. $\endgroup$ – Pedro Tamaroff Aug 16 '12 at 21:52
  • $\begingroup$ @AndreyRekalo can you explain more why that the first integral at the right-hand side tends to zero by the Riemann-Lebesgue lemma ? I read the article at wiki but i faild to connect between the integral and law of Riemann–Lebesgue lemma $\endgroup$ – mhd.math Aug 15 '13 at 8:01
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Choose any $\gamma$ such that $0 < \gamma < 1/2$ and let $c = \sin\gamma$. Then intervals of $x$ in which $|\sin x|< c$ have length $2\gamma < 1$, and intervals in which $|\sin x| \ge c$ have length $\pi - 2\gamma > 1$. Hence if $|\sin x| < c$ then $|\sin(x + 1)| \ge c$. Let $n_1 < n_2 < n_3 \cdots$ be the values of $n$ for which $|\sin n| \ge c$. Then $n_1 = 1$, and by the above $n_{k+1} = n_k + 1$ or $n_k + 2$. So for all $k$ we have $n_k < 2k$ and hence $(\sin^2 n_k)/n_k > c^2/2k$. Since $1/1 + 1/2 + 1/3 + \cdots$ is divergent, the result follows.

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    $\begingroup$ Isn't this an answer to a different question? The question is about $\sum \frac{1}{k} \sin kx$ (which is convergent), not $\sum \frac{1}{k} \sin^2 kx$. $\endgroup$ – epimorphic Jan 27 at 18:41

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