Confusion regarding infinite series involving complex number I've been asked to find $\Omega$, such that :
$$\Omega=\frac{\Omega_1}{\Omega_2}=\frac{1+e^{\frac{2i\pi}{3}}+e^{\frac{4i\pi}{3}}+e^{\frac{6i\pi}{3}}+e^{\frac{8i\pi}{3}}+.....}{i+\frac{i^2}{2}+\frac{i^3}{4}+\frac{i^4}{8}+.......}$$
I noticed that in the numerator, $e^{\frac{2i\pi}{3}}=(e^{2\pi i})^{1/3}=1^{1/3}=\omega$ which is the cube root of unity.
Hence, $e^{\frac{4i\pi}{3}}=(e^{2\pi i})^{2/3}=1^{2/3}=\omega^2$.
Similarly, $e^{\frac{6i\pi}{3}}=(e^{2\pi i})^{3/3}=1^{3/3}=\omega^3=1$, and the pattern repeats.
Since, $1+\omega+\omega^2=0$, The entire series $\Omega_1$ must also be equal to $0$, and hence $\Omega=\frac{\Omega_1}{\Omega_2}=0$
However, this is what one of my teachers did in a note :


I don't see how this second method is correct, and my method is wrong. I don't think that in the second method, we can add the successive terms in the gp series like that.
Any help would be highly appreciated.
 A: The partial sums of the numerator are a non-constant periodic sequence $\,(1, 1+\omega,0)\,$, so $\,\Omega_1\,$ does not converge, and therefore both answers are wrong.

Since, $1+\omega+\omega^2=0$, the entire series $\Omega_1$ must also be equal to $0$

Grouping terms is only allowed if the series is absolutely convergent, which is not the case for $\,\Omega_1\,$, see for example here.

(infinite GP series) $\displaystyle \quad \dots = \frac{1}{1 - e^{\frac{2\pi i}{3}}}$

The GP series formula is only valid for convergent series $\,\sum z^k\,$ with $\,|z| \lt 1\,$. It cannot be applied in this case since $\,|z|=1\,$ and the series diverges, see for example here.
A: Consider $z=-1 + 1 + -1 +1 + -1 +1+...$. Well, it alternates, so it doesn't converge. It's also ambiguous.
Does $z=(-1+1)+(-1+1)+...=0+0+0=0$
How about $z=-1+(1+-1)+(1+-1)+...=-1+0+0+0=-1$? Swap adjacent terms pair-wise and the arguments says $z=1$.
Yet another argument says $\frac{1}{1-z}=1+z+z^2+...$ so $\frac{1}{1-(-1)}=\frac{1}{2}=1+-1+1+-1+...$
$\Omega_1=\frac{1}{1-e^{\frac{2\pi i}{3}}}$
$\Omega_2=\frac{i}{1-i/2}$
So $\frac{\Omega_1}{\Omega_2}=\frac{1-i/2}{i(1-e^{2\pi i/3})}=\frac{(1-i/2)(-i)(1-e^{-2\pi i/3})}{(1-e^{2\pi i/3})(1-e^{-2\pi i/3})}$ , rationalizing the denominator.
$e^{2\pi i/3}=-1/2+i(\sqrt{3}/2)$
$1-e^{2\pi i /3}=3/2-i(\sqrt{3}/2)$
$(1-e^{2\pi i /3})(1-e^{-2\pi i /3})=3$
$-(1/2+i)(3/2+i\sqrt{3}/2)=-(3-2\sqrt{3})/4-i(6+\sqrt{3})/4$
So $\frac{\Omega_1}{\Omega_2}=\frac{(2\sqrt{3}-3)-i(6+\sqrt 3))}{12}$ if you use this series approach.
Take that with a grain of salt though. There are doubtless other ways to associate a value with it. Analytic Continuation should probably be mentioned, but the details are vague .
