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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
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    $\begingroup$ @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 $\endgroup$ – mnsh Aug 15 '13 at 8:01
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For any $n\geq 1$, we know where the stationary points of $F_n(x)$ occur since $F_n'(x)$ has a simple closed form.
It follows that $$ \sup_{x\in\mathbb{R}} |F_n(x)| = \sum_{k=1}^{n}\frac{1}{k}\sin\left(\frac{2\pi k}{2n+1}\right) $$ and over $[0,\pi]$ we have $\sin(x)\leq \frac{4}{\pi^2}x(\pi-x)$ by concavity, therefore $$ \sup_{x\in\mathbb{R}} |F_n(x)| \leq \frac{8n^2}{(2n+1)^2} <2.$$ We may also prove that the sequence $$ A_n = \sum_{k=1}^{n}\frac{1}{k}\sin\left(\frac{2\pi k}{2n+1}\right) $$ is increasing and convergent to $$ \int_{0}^{\pi}\frac{\sin x}{x}\,dx = \text{Si}(\pi) \approx 1.85194. $$ Ultimately we may check that $\sum_{k\geq 1}\frac{\sin(kx)}{k}$ is the Fourier series of the sawtooth wave, i.e. the $2\pi$-periodic extension of $\frac{\pi-x}{2}$ defined over $(0,2\pi)$. This is enough to ensure convergence in $L^2$, plus we have a uniform bound for $|F_n(x)|$.

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