By application of Euler Identities we get that the Fourier Series of $f$ which is given by: $$f(\theta)\thicksim \sum_{n \ne o}\frac{(-1) ^ {n + 1}}{in} e^{in\theta} $$ is equal to $$2\sum_{n=1}^{\infty}(-1)^{n+1}\frac{sinn\theta}{n} $$
I have divided the summation $\sum_{n\ne0}$ to $\sum_{n=1}^{\infty}\frac{(-1) ^ {n + 1}}{in} e^{in\theta} $ and $\sum_{n=1}^{\infty}\frac{(-1) ^ {-n + 1}}{i(-n)} e^{i(-n)\theta} $ and I know that in the second summation $ \frac{(-1) ^ {-n + 1}}{i(-n)}$ will become $ \frac{(-1) ^ {-n + 2}}{i(n)}$, So How can I reach the final formula that he wants?
Hint: $$f(\theta)\thicksim \sum_{n \ne o}\frac{(-1) ^ {n + 1}}{in} e^{in\theta} =\sum_{n < o}\frac{(-1) ^ {n + 1}}{in} e^{in\theta}+\sum_{n > o}\frac{(-1) ^ {n + 1}}{in} e^{in\theta}=\sum_{n > o}\frac{(-1) ^ {-n + 1}}{-in} e^{-in\theta}+\sum_{n > o}\frac{(-1) ^ {n + 1}}{in} e^{in\theta}=\sum_{n > o}\frac{(-1) ^ {n + 1}}{in} (-e^{-in\theta}+e^{in\theta})$$