4

How to prove the following series converge for every $\theta$, any suggestions are welcome?

$$ \sum_{n=1}^{\infty}\frac{(-1)^n\sin(n\theta)}{n}.$$

Mhenni Benghorbal
  • 47,431
Wu Yufeng
  • 121

2 Answers2

2

Your series is just the Fourier sine series of the (sawtooth wave) periodic continuation of the function $f(\theta)=-\frac{\theta}{2}$ defined over $[-\pi,\pi]$, hence: $$\sum_{n=1}^{+\infty}\frac{(-1)^n \sin(n\theta)}{n}=\left\{\begin{array}{rcl}-\frac{1}{2}(\theta\operatorname{mod} 2\pi)&\text{if}&\theta\not\in\pi\mathbb{Z},\\ 0&\text{if}&n\in\pi\mathbb{Z},\end{array}\right.$$ where $x\to x\operatorname{mod} 2\pi$ is intended to map any real number into $[-\pi,\pi]$. This just follows from: $$-\frac{1}{\pi}\int_{-\pi}^{+\pi}\frac{x}{2}\sin(nx)\,dx = \frac{(-1)^n}{n}$$ that can be checked through integration by parts.

As an alternative, you can prove convergence by noticing that: $$\sum_{n=1}^N \frac{(-1)^n\sin(n\theta)}{n}=\sum_{n=1}^N \frac{\cos(\pi n)\sin(n\theta)}{n}=\frac{1}{2}\left(\sum_{n=1}^N \frac{\sin((\theta+\pi)n)}{n}+\sum_{n=1}^N \frac{\sin((\theta-\pi)n)}{n}\right)$$ and since the partial sums of $\sin(\psi n)$, for any $\psi\not\in\pi\mathbb{Z}$, are bounded by $\frac{1}{|\sin(\psi/2)|}$, the conditions for Dirichlet's test are met.

Jack D'Aurizio
  • 353,855
  • 1
    Thank you so much for your hint, I solved it by using Dirichlet's test. – Wu Yufeng Sep 08 '14 at 07:08
  • Would you please briefly explain me the logic of your last equality? Moreover, why the partial sums of $\sin(\psi n)$ are bounded by $1/|\sin(\psi /2)|$? – Charlie Sep 08 '14 at 10:02
  • 1
    The last equality just follows from Briggs' formulas. To prove that the partial sums of $\sin(\psi n)$, consider that the sine function is the imaginary part of the complex exponential. Since $\sum_{n\in I} e^{i\psi n}$ is the sum of a geometric sequence, the bound just follows from factoring $\Im\sum_{n\in I} e^{i\psi n}$ in terms of trigonometric functions. – Jack D'Aurizio Sep 08 '14 at 10:16
1

Hint: You can use Dirichlet's test.

Added: You need to bound partial sums

$$ \sum_{i}^{n} (-1)^i\sin(i\theta) = {\frac { \left( -1 \right) ^{n}\sin \left( \theta\,n \right) + \left( -1 \right) ^{n}\sin \left( \theta\,n+\theta \right) -\sin \left( \theta \right) }{2+2\,\cos \left( \theta \right) }}$$

which is easy to do.

Mhenni Benghorbal
  • 47,431