Is $\sum_{n=1}^{\infty}{\frac{\sin(nx)}n}$ continuous?

Question

Considering the infinite series $\sum_{n=1}^{\infty}{\frac{\sin(nx)}n}$ , I can show that it is not convergent uniformly by Cauchy's criterion and that it is convergent for every $x$ by Dirichlet's test. But I don't know how to judge whether it is continuous.

Could you tell me the answer and why? Thank you in advance!

How do you show convergence by Dirichlet's test? For example if $x=1$ it isn't alternately positive then negative... — coffeemath, Nov 14 '13 at 15:11
$1/n$ is decreasing and the partial sum of $\sum_{n=1}^{\infty}{sin(nx)}$ is bounded. So the sum of the product is convergent. — FFGG, Nov 14 '13 at 15:14
Is it somehow obvious that the partial sums of $\sum \sin(nx)$ are bounded? I'm not seeing it... — Jason DeVito - on hiatus, Nov 14 '13 at 15:58
@JasonDeVito It is the imaginary part of a geometric series. — achille hui, Nov 14 '13 at 16:00

score 18 · Accepted Answer · answered Nov 14 '13 at 15:40

18

To inspect the discontinuity of the summation, let's calculate the sum. By the Abel's theorem,

$$ f(x) := \sum_{n=1}^{\infty} \frac{\sin nx}{n} = \lim_{s\to 0^{+}} \sum_{n=1}^{\infty} \frac{\sin nx}{n} e^{-ns}. $$

By utilizing Taylor expansion of the logarithm,

\begin{align*} \sum_{n=1}^{\infty} \frac{\sin nx}{n} e^{-ns} &= \Im \sum_{n=1}^{\infty} \frac{e^{n(ix-s)}}{n} = - \Im \log (1 - e^{ix-s}) \\ &= -\Im \log (1 - e^{-s}\cos x - ie^{-s}\sin x) \\ &= \arctan \left(\frac{e^{-s}\sin x}{1 - e^{-s}\cos x}\right). \end{align*}

Thus taking $s \to 0^{+},$

$$ f(x) = \arctan \left(\frac{\sin x}{1 - \cos x}\right) = \arctan \left(\cot \frac{x}{2}\right) = \arctan \left(\tan \frac{\pi-x}{2}\right). $$

Therefore

$$ f(x) = \begin{cases} \frac{\pi - x}{2} & x \in (0, 2\pi),\\ 0 & x = 0, \\ f(x+2\pi), & x \in \Bbb{R}. \end{cases} $$

This shows a clear-cut jump discontinuity at each $x \in 2\pi \Bbb{Z}$.

answered Nov 14 '13 at 15:40

Sangchul Lee

167,468

2

I rethink this question today.And I observe that if we drop out "$e^{-ns}$",the proof can still work,in other words, for $x\neq0$ \begin{align} \sum_{n=1}^{\infty} \frac{\sin nx}{n} &= \Im \sum_{n=1}^{\infty} \frac{e^{nix}}{n} = - \Im \log (1 - e^{ix}) \ &= -\Im \log (1 - \cos x - i\sin x) \ &= \arctan \left(\frac{\sin x}{1 - \cos x}\right). \end{align}
And if $x=0$ ,the summation is $0$. – FFGG Nov 28 '13 at 08:01
2

@F.G Tha ks for pointing out that. I was also aware of that, but I adopted this regularizing method in order to avoid possible problems arising from the boundary behavior and the convergence mode of the series. – Sangchul Lee Nov 28 '13 at 08:08
What is that J symbol? – william_grisaitis Nov 24 '21 at 06:41
The "J" symbol is a fancy I, and it means the imaginary part – Theone Dec 19 '21 at 22:21

score 1 · Answer 2 · answered Sep 21 '19 at 17:15

Another way to prove that $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is discontinuous at the origin (and so at every element of $2\pi\mathbb{Z}$).
If $f(x)$ is a bounded function and $\lim_{x\to 0^+}f(x)$ does exist, it equals $\lim_{m\to +\infty}\int_{0}^{+\infty} f(x)m e^{-mx}\,dx $ (approximation of the identity). Due to $$ \int_{0}^{+\infty}\sin(nx)m e^{-mx}\,dx = \frac{m n}{m^2+n^2} $$ we have $$ \lim_{x\to 0^+}\sum_{n\geq 1}\frac{\sin(nx)}{n} = \lim_{m\to +\infty}\sum_{n\geq 1}\frac{m}{m^2+n^2} $$ and by Riemann sums the RHS of the last line equals $\int_{0}^{+\infty}\frac{dx}{x^2+1}=\frac{\pi}{2}\neq 0$.

Is $\sum_{n=1}^{\infty}{\frac{\sin(nx)}n}$ continuous?

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