Concerning the sum $\sum_{n = 1}^\infty \sin nx$

Question

I recently came across this question and I posted an answer. It has been pointed out that my answer is incorrect. I cannot work out what is wrong with my reasoning. The answer I gave corresponds with the Abel and Cesaro sum, so perhaps $\sum$ is not the usual summation operator? Am I correct in asserting that if $x$ is in the upper half-plane, i.e., $\textbf{I}[x] > 0$, then $|e^{ix}| < 1$ and consequently $$\sum_{n = 1}^\infty e^{inx} = \frac{e^{ix}}{1 - e^{ix}},$$ or is my argument flawed? Any help would be appreciated.

Answer 1

Because you assume that $x$ is not real, the imaginary part of $e^{inx}$ is not $\sin nx$.

Also, when computing the conjugate if $1-e^{ix}$, you don't get $1-e^{-ix}$ when $x$ is non-real, but rather $$1-e^{-i\bar x}$$