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I want to prove $2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx) $ converges pointwise and uniformly to $x$ on $[-\pi,\pi]$. I know $\sum_{k=1}^{\infty}\frac{(-1)^n}{n}$ converge by alternating series test. And $\sum a_n \sin(nx)$ converge by Dirichlet test if $a_n$ is decreasing sequence. But in this case this does not work. Maybe it just we can just consider the interval without $-\pi$,$\pi$. I get lost. Please help. Thanks a lot After trying, I think maybe there is no uniform convergence?

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  • $\begingroup$ The proof in Fourier Analysis is quite elementary. It uses the simple idea of 'summation by parts'. $\endgroup$ – Kavi Rama Murthy Apr 28 at 12:04
  • $\begingroup$ @KaviRamaMurthy. Would you mind pointing me to the right book?or would you illustrate it? Thanks a lot. $\endgroup$ – user780338 Apr 28 at 12:11
  • $\begingroup$ The book by Edwards on Fourier series has a section called 'The series (C) and (S) as Fourier series'. There is nice discussion of series of the type $\sum a_n \sin (nx)$ and $\sum a_n \cos (nx)$. $\endgroup$ – Kavi Rama Murthy Apr 28 at 12:15
  • $\begingroup$ Well it may converge point-wise and uniformly, but in which interval ? $\endgroup$ – Siddhartha Apr 28 at 12:57
  • $\begingroup$ @Siddhartha I don't know for sure. Does (-pi,pi) work? Please guide me $\endgroup$ – user780338 Apr 28 at 13:21
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Convergence is not uniform on $(-\pi,\pi)$ (although it is on compact subintervals).

To prove non-uniform convergence note that

$$2\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin nx }{n} = -2\sum_{n=1}^{\infty} \cos n\pi \frac{\sin nx }{n} = -2 \sum_{n=1}^{\infty} \frac{\sin n(\pi+x) }{n} $$

However, taking $x_n = -\pi + \frac{\pi}{4n} \in (-\pi,\pi)$ we have for $n < k \leqslant 2n$ that $\frac{\pi}{4} < k (\pi+x_n) \leqslant \frac{\pi}{2}$ which implies $\frac{1}{\sqrt{2}} < \sin k (\pi+x_n) \leqslant 1$ and for all $n \in \mathbb{N}$,

$$\sup_{x \in (-\pi,\pi)}\left| \sum_{k = n+1}^{2n}\frac{\sin k(\pi+x) }{k} \right|\geqslant \sum_{k = n+1}^{2n}\frac{\sin k(\pi+x_n) }{k} > \frac{1}{\sqrt{2}}\ \sum_{k=n+1}^{2n} \frac{1}{k} > \frac{1}{\sqrt{2}} \cdot n \cdot \frac{1}{2n} = \frac{1}{2\sqrt{2}}$$

The LHS fails to converge to $0$ as $n \to \infty$ and the Cauchy criterion for uniform convergence is violated.

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  • $\begingroup$ Nicely done! I hope that you're doing well my friend. $\endgroup$ – Mark Viola Apr 29 at 7:04
  • $\begingroup$ The OP originally asked to show UC. Since the, the question was edited. So, I've added a new section with an alternative proof to yours. I'd be grateful if you would have a look whenever you have a moment. ;-) $\endgroup$ – Mark Viola Apr 29 at 20:32
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NOTE: The original question that the OP asked was

"Prove $2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx) $ converge pointwise and uniformly to $x$ on $[0,2\pi]$ using elementary analysis"**


Let $a_n(x)=(-1)^{n-1}\sin(nx)$ and $b_n(x)=\frac1n$. Obviously, $b_n(x)\to 0$ monotonically and uniformly as $n\to\infty$.

Moreover, for any $0<\delta_1<\pi$ and $0<\delta_2<\pi$, and $x\in [-\pi+\delta_1,\pi-\delta_2]$,

$$\begin{align} \left|\sum_{n=1}^N a_n(x)\right|&=\left|\sum_{n=1}^N (-1)^{n-1}\sin(nx)\right|\\\\ &\le\left|\sec(x/2)\right|\\\\ &\le \max(\csc(\delta_1),\csc(\delta_2)) \end{align}$$

Therefore, Dirichlet's Test guarantees that the series $\sum_{n=1}^\infty \frac{(-1)^{n-1}\sin(nx)}{n}$ converges uniformly on $[-\pi+\delta_1,\pi-\delta_2]$.


EDITED: After the OP changed the question

We now give a proof that the series $2\sum_{n=1}^\infty \frac{(-1)^{n-1}\sin(nx)}{n}$ fails to converge uniformly for $x\in (-\pi,\pi)$.

We first note that the series converges to $-x$ for $x\in (-\pi,\pi)$. That is to say that the Fourier series for $x$ on $(-\pi,\pi)$ is given by

$$x=2\sum_{n=1}^\infty \frac{(-1)^{n-1}\sin(nx)}{n}$$

Now let $f_N(x)$ be the $N$th partial sum of the Fourier series for $x$. Then, denoting $t=x+\pi$ we can write

$$\begin{align} f_N(x)&=2\sum_{n=1}^N\frac{(-1)^{n-1}\sin(nx)}{n}\\\\ &=-2\sum_{n=1}^N \frac{\sin(nt)}{n}\\\\ &=-2\int_0^t \sum_{n=1}^N \cos(nu)\,du\\\\ &=t-\int_0^t \frac{\sin((N+1/2)u)}{\sin(u/2)}\,du\\\\ &=t-\int_0^{(N+1/2)t}\frac{\sin(x)}{x}\frac{x/(2N+1)}{\sin(x/(2N+1))}\,dx \end{align}$$

It suffices to show that $\int_0^t \frac{\sin((N+1/2)u)}{\sin(u/2)}\,du$ fails to converge uniformly to $\frac\pi2$ for $t\in (0,2\pi)$. Now take $t=1/(N+1/2)$

Then, we see that

$$\sin(1)\le\int_0^1 \frac{\sin(x)}{x}\frac{x/(2N+1)}{\sin(x/(2N+1))}\,dx\le \csc(1)$$

Hence we conclude that the convergence of $f_N(x)$ fails to converge uniformly on $(-\pi,\pi)$. And we are done!

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  • $\begingroup$ Could you explain more about the two inequality sign? Thank you so much $\endgroup$ – user780338 Apr 29 at 5:07
  • $\begingroup$ @user123469123131 The sum $\sum_{n=1}^N (-1)^{n-1}\sin(nx)$ can be evaluated in closed form from which the bound is easily seen. In the second, the secant function $\sec(x/2)$ is a maximum at $x=\pi-\delta$. $\endgroup$ – Mark Viola Apr 29 at 5:34
  • $\begingroup$ do you proof hold when I replace 0 in the proof by $-\pi$?Cuz I want to prove uniform convergence on $(-pi,pi)$ now. Thanks a lot $\endgroup$ – user780338 Apr 29 at 5:45
  • $\begingroup$ @ Mark Viola Is you proof still correct when I replace 0 in the proof by $-\pi$?Cuz I want to prove uniform convergence on $(-\pi,\pi)$ now. Please examine my edit to see if it is OK.Thanks a lot $\endgroup$ – user780338 Apr 29 at 5:51
  • $\begingroup$ Yes. For any $\delta_1>0$ and $\delta_2>0$, the convergence is uniform on $[-\pi+\delta_1,\pi-\delta_2]$. $\endgroup$ – Mark Viola Apr 29 at 6:57

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