# Prove or Disprove $2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx)$ converge uniformly to $x$ on $(-\pi,\pi)$

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?

• The proof in Fourier Analysis is quite elementary. It uses the simple idea of 'summation by parts'. – Kavi Rama Murthy Apr 28 at 12:04
• @KaviRamaMurthy. Would you mind pointing me to the right book?or would you illustrate it? Thanks a lot. – user780338 Apr 28 at 12:11
• 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 $\sum a_n \cos (nx)$. – Kavi Rama Murthy Apr 28 at 12:15
• Well it may converge point-wise and uniformly, but in which interval ? – Siddhartha Apr 28 at 12:57
• @Siddhartha I don't know for sure. Does (-pi,pi) work? Please guide me – user780338 Apr 28 at 13:21

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.



• Nicely done! I hope that you're doing well my friend. – Mark Viola Apr 29 at 7:04
• 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. ;-) – Mark Viola Apr 29 at 20:32

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!

• Could you explain more about the two inequality sign? Thank you so much – user780338 Apr 29 at 5:07
• @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 $x=\pi-\delta$. – Mark Viola Apr 29 at 5:34
• 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 – user780338 Apr 29 at 5:45
• @ 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 – user780338 Apr 29 at 5:51
• Yes. For any $\delta_1>0$ and $\delta_2>0$, the convergence is uniform on $[-\pi+\delta_1,\pi-\delta_2]$. – Mark Viola Apr 29 at 6:57