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!

  • $\begingroup$ How do you show convergence by Dirichlet's test? For example if $x=1$ it isn't alternately positive then negative... $\endgroup$
    – coffeemath
    Nov 14 '13 at 15:11
  • 4
    $\begingroup$ $1/n$ is decreasing and the partial sum of $\sum_{n=1}^{\infty}{sin(nx)}$ is bounded. So the sum of the product is convergent. $\endgroup$
    – F.G.
    Nov 14 '13 at 15:14
  • $\begingroup$ Yes, got it. Thanks, and +1 for an interesting question. $\endgroup$
    – coffeemath
    Nov 14 '13 at 15:16
  • $\begingroup$ Is it somehow obvious that the partial sums of $\sum \sin(nx)$ are bounded? I'm not seeing it... $\endgroup$ Nov 14 '13 at 15:58
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    $\begingroup$ @JasonDeVito It is the imaginary part of a geometric series. $\endgroup$ Nov 14 '13 at 16:00

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 $s \to 0^{+},$

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


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

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

  • 2
    $\begingroup$ 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 $0$. $\endgroup$
    – F.G.
    Nov 28 '13 at 8:01
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    $\begingroup$ @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. $\endgroup$ Nov 28 '13 at 8:08

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 $\int_{0}^{+\infty}\frac{dx}{x^2+1}=\frac{\pi}{2}\neq 0$.


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