Showing $\sum\frac{\sin(nx)}{n}$ converges pointwise

Question

I do not understand how one can say using "Dirichlet conditions" that $\sum_{n=1}^{\infty}\dfrac{\sin(nx)}{n}$ is pointwise convergent. I know the proof for $x=1$ but how can one show it is convergent for all $x>0$, pointwise? The Dirichlet theorem for series of functions only lays down criteria for uniform convergence of series, not pointwise convergence. One method to show the series is convergent pointwise is to show it is the Fourier sum of the $2\pi$ periodic function $\dfrac{\pi-x}{2}$ in (which is differentiable and hence the Fourier sum converges to the function) but I want to prove it without using Fourier series.

Please note that I have gone through all the sites offered by MSE where this question has been posted and in almost all of them, upvoted answers just write down "USING DIRICHLET CONDITIONS" and then possibly give a link to the Wiki page. However, I do not understand how one can use Dirichlet theorem to prove just the pointwise convergence of a series.

Any help is appreciated.

Answer 1

Dirichlet Test: Let $(a_n)$ and $(b_n)$ be two sequences satisfying

1. $\lim_{n\to+\infty}a_n=0$,
2. $\sum_{n=1}^{+\infty}|a_{n+1}-a_{n}|$ converges, and
3. There exists a number $M$ such that $\bigl|\sum_{n=1}^m b_n\bigr|\leq M$ for all $m\geq 1$.

Then the sum $\sum_{n=1}^{+\infty} a_nb_n$ converges.

Let us apply that to the sum $\sum_{n=1}^{+\infty}\frac{\sin (nx)}{n}$. I will leave some details to you to fill in.

To prove that the sum is pointwise convergent, we fix $x\in (0,2\pi)$ (The series is clearly $2\pi$ periodic, so we can reduce to $[0,2\pi]$, but the endpoints are trivial).

Let $a_n=1/n$ and $b_n=\sin (nx)$. Then the first two conditions are satisfied (the second since $a_{n+1}-a_n=\frac{1}{n(n+1)}$, and the series with terms $1/(n(n+1))$ is absolutely convergent).

The third condition follows from the well-known formula $$ \sum_{n=1}^m \sin(nx)=\frac{\sin(mx/2)\sin((m+1)x/2)}{\sin (x/2)}, $$ since then we have $$ \Bigl|\sum_{n=1}^m \sin(nx)\Bigr|\leq \frac{1}{\sin (x/2)} $$ independently of $m$. Thus $\sum_{n=1}^{+\infty}\frac{\sin (nx)}{n}$ converges.

Answer 2

We can use the Abel summation and get$$\sum_{n\leq N}\frac{\sin\left(nx\right)}{n}=H_{N}\sin\left(Nx\right)-x\int_{1}^{N}H_{\left\lfloor t\right\rfloor }\cos\left(tx\right)dt$$ and now using the asymptotic $$H_{n}=\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right)$$ we have $$\sum_{n\leq N}\frac{\sin\left(nx\right)}{n}=\left(\log\left(N\right)+\gamma\right)\sin\left(Nx\right)-x\int_{1}^{N}\left(\log\left(t\right)+\gamma\right)\cos\left(tx\right)dt+O\left(\frac{\sin\left(Nx\right)}{N}\right)+O\left(x\int_{1}^{N}\frac{\cos\left(tx\right)}{t}dt\right).$$ Now if we integrate by parts the first integral we get $$x\int_{1}^{N}\left(\log\left(t\right)+\gamma\right)\cos\left(tx\right)dt=\left.\left(\log\left(t\right)+\gamma\right)\sin\left(tx\right)\right|_{1}^{N}-\int_{1}^{N}\frac{\sin\left(tx\right)}{t}dt$$ and so $$\sum_{n\leq N}\frac{\sin\left(nx\right)}{n}=\gamma\sin\left(x\right)+\int_{1}^{N}\frac{\sin\left(tx\right)}{t}dt+O\left(\frac{\sin\left(Nx\right)}{N}\right)+O\left(x\int_{1}^{N}\frac{\cos\left(tx\right)}{t}dt\right).$$ Now if we take $N\rightarrow\infty$ $$\sum_{n\geq1}\frac{\sin\left(nx\right)}{n}=\gamma\sin\left(x\right)+\int_{1}^{\infty}\frac{\sin\left(tx\right)}{t}dt+O\left(x\int_{1}^{\infty}\frac{\cos\left(tx\right)}{t}dt\right)$$ and we have $$\int_{1}^{\infty}\frac{\sin\left(tx\right)}{t}dt=\int_{x}^{\infty}\frac{\sin\left(u\right)}{u}du=-\textrm{Si}\left(x\right)$$ where $\textrm{Si}\left(x\right)$ is the sine integral. Similar way for the integral with the cosine, so we have pointwise convergence.

Answer 3

This problem is in Fourier Analysis an Introduction Chapter 3 Exercise 19. and 20.

19.Give another prove that the sum $\sum_{0<|n|\le N}e^{inx}/n$ is uniformly bounded in N and $x \in [-\pi,\pi]$ by using the fact that

$$\frac{1}{2i}\sum_{0<|n|\le N}\frac{e^{inx}}{n}=\sum_{n=0}^{N}\frac{\sin{n}}{n}=\frac{1}{2}\int_{0}^{x}(D_{N}(t)-1)dt$$

20. Let $f(x)$ denote the sawtooth function defined by $f(x)=(\pi-x)/2$ on the interval $(0,2\pi)$ with $f(0)=0$ and extended by periodicity to all of $\mathcal R.$ The Fourier series of $f$ is $$f(x) \sim \frac{1}{2i}\sum_{\vert n \vert ≠0}\frac{e^{inx}}{n}=\sum_{n=1}^{\infty}\frac{\sin{nx}}{n},$$ and $f$ has a jump discontinuity at the origin with $$f(0^{+})=\frac{\pi}{2},\ f(0^{-})=-\frac{\pi}{2},$$ and hence $$f(0^{+})-f(0^{-})=\pi.$$ Show that $$\max_{0<x\le \pi/N}S_{N}(f)(x)-\frac{\pi}{2}=\int_{0}^{\pi}\frac{\sin(t)}{t}dt -\frac{\pi}{2},$$ because $f(x)$ is a periodic function, we just need to consider the convergence in $[-\pi,\pi],$ and $$\sum_{n=0}^{\infty}\frac{\sin{nx}}{n}=\lim_{N\to \infty}\sum_{n=0}^{N}\frac{\sin{nx}}{n}$$ we just need to show that $$\lim_{N \to \infty}\max_{0<x\le \pi} S_{N}(f)(x)<\infty$$ The detailed prove is in this link.