# 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.

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.

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 .