Calculating series using residue theorem: $ \sum_{n=0}^{\infty} \left(\frac{(-1)^n}{2^n\cdot n}\right) \sin(n \theta) $

Question

I don't know how to approach this series. Assuming that it is done using the residue theorem, should the answer be 0? Because the function within doesn't have any pole.

$$ \sum_{n=0}^{\infty} \left(\frac{(-1)^n}{2^n\cdot n}\right) \sin(n \theta) $$

Thanks in advance for your help.

Answer 1

I see no obvious way to apply the residue theorem.

Hint: Note that $$ \sum_{n=1}^{\infty} \left(\frac{(-1)^n}{2^n\cdot n}\right) \sin(n \theta) = \operatorname{Im}\left[ \sum_{n=1}^{\infty} \frac{\left(-\frac12 e^{i\theta}\right)^n}{n} \right] $$

Answer 2

The result is $$\theta-\tan^{-1}\left(\frac{\sin\theta}{2+\cos\theta}\right)$$

I will add a derivation later.

Derivation:

Let $S(\theta)$ be your sum.

Note that your sum converges uniformly. Now, differentiate term-by-term with respect to $\theta$.

$$S’=\sum^\infty_{n=0}\left(\frac{-1}2\right)^n\cos(n\theta)=\Re\sum^\infty_{n=0}\left(\frac{-e^{i\theta}}2\right)^n=\Re\frac1{1+\frac{e^{i\theta}}2}$$

Integrating gives $$S=\Re(\theta+i\ln(2+e^{i\theta}))+C$$

Since $$\Im \ln(2+e^{i\theta})=\text{arg}(2+e^{i\theta})=\tan^{-1}\left(\frac{\sin\theta}{2+\cos\theta}\right)$$

Thus, $$S=\theta-\tan^{-1}\left(\frac{\sin\theta}{2+\cos\theta}\right)+C$$

By considering $S(0)=0$, we get $C=0$.

So, $$\color{red}{S=\theta-\tan^{-1}\left(\frac{\sin\theta}{2+\cos\theta}\right)}$$