Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
2  
$1/n$ is decreasing and the partial sum of $\sum_{n=1}^{\infty}{sin(nx)}$ is bounded. So the sum of the product is convergent. –  F.G Nov 14 '13 at 15:14
    
Yes, got it. Thanks, and +1 for an interesting question. –  coffeemath Nov 14 '13 at 15:16
    
Is it somehow obvious that the partial sums of $\sum \sin(nx)$ are bounded? I'm not seeing it... –  Jason DeVito Nov 14 '13 at 15:58
    
@JasonDeVito It is the imaginary part of a geometric series. –  achille hui Nov 14 '13 at 16:00

1 Answer 1

up vote 10 down vote accepted

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 $x \in 2\pi \Bbb{Z}$.

share|improve this answer
    
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 $0$. –  F.G Nov 28 '13 at 8:01
1  
@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. –  sos440 Nov 28 '13 at 8:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.