Show that $\sum_{n \neq 0} \frac{(-1)^{n+1}}{in} e^{in\theta} = 2 \sum_{n=1}^\infty (-1)^{n+1} \frac{\sin n\theta}{n}$.

Question

Show that $$\sum_{n \neq 0} \frac{(-1)^{n+1}}{in} e^{in\theta} = 2 \sum_{n=1}^\infty (-1)^{n+1} \frac{\sin n\theta}{n}.$$

This is not an exercise. It is an example from Stein, Fourier Analysis - An Introduction, pg. 36. I'm rusty on my complex algebra and I'm having trouble understanding why this equality is true. I used the complex exponential identity $e^{ix} = \cos(x) + i\sin(x)$ without success.

Answer 1

The key here is that $e^{in\theta} = \cos(n\theta) + i\sin(n\theta)$ implies that $$ e^{in\theta} - e^{-in\theta} = 2i\sin(n\theta). $$ Thus the sum on the left hand side is, after rearranging terms (grouping those indexed by $\pm n$) and simplifying, $$ \sum_{n > 0} \frac{(-1)^{n+1}}{in}(e^{in\theta}-e^{-in\theta}) = 2 \sum_{n > 0} (-1)^{n+1} \frac{\sin(n\theta)}{n}. $$ I'll leave the question to you of why this reordering is acceptable here.

Answer 2

Use $\sin n\theta = \dfrac{e^{in\theta} - e^{-in\theta}}{2i}$ and observe the sum on the left can be written as $$ \sum_{n=1}^\infty \left[ \frac{(-1)^{n+1}}{in} e^{in\theta} + \frac{(-1)^{-n+1}}{i(-n)} e^{i(-n)\theta} \right].$$

Answer 3

By direct calculation, $\sum_{n\neq0}\frac{\left(-1\right)^{n+1}}{in}e^{in\theta} =\sum_{n=-\infty}^{-1}\frac{\left(-1\right)^{n+1}}{in}e^{in\theta}+\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{in}e^{in\theta} =\sum_{n=1}^{\infty}\frac{\left(-1\right)^{-n+1}}{\left(-1\right)in}e^{-in\theta}+\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{in}e^{in\theta} =\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{in}\left(e^{in\theta}-e^{-in\theta}\right) =2\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\frac{\sin\left(n\theta\right)}{n}.$