I was thinking and revisiting a problem that I had to solve last year in an Introductory Complex Analysis exam. It went like this: Present an example of a convergent series of complex numbers $\sum_{n=1}^{\infty}z_n$ such that $\sum_{n=1}^{\infty}z_n^3$ diverges.
My solution was to make use of the cube root of unity, by defining $\large z_n=\frac{e^{2\pi i n/3}}{\sqrt[3]{n}}$. In this fashion, one can easily see that $\large z_n^3=\frac{e^{2\pi i n}}{n}=\frac{1}{n}$, and so $\sum_{n=1}^{\infty}z_n^3$ diverges by the harmonic series. However, I was not able to prove that the series of $z_n$ was convergent. I am somewhat positive that the series is indeed convergent, as WolframAlpha computes that it converges to $\text{Li}_{1/3}(e^{2\pi i/3})$. It's quite obvious that the convergence won't be absolute, as $|z_n|=\frac{1}{\sqrt[3]{n}}$ gives a divergent series. I attempted to separate $z_n=a_n+b_n i$ for $a_n=\cos(2\pi n/3)/\sqrt[3]{n}$ and $b_n=\sin(2\pi n/3)/\sqrt[3]{n}$, but I couldn't get this to be very fruitful. Any ideas?
Note: I also know that there are sequences $z_n$ that satisfy this property and are such that $\mathfrak{Im}(z_n)=0$, but this is not what I'm looking for.
Best Answer
The convergence of the series can be proven as follows: first of all you have the characterisation: a complex series $\sum_{n=1}^\infty z_n$, with $z_n=x_n+i y_n$ converges to $z_*=x_*+iy_*$ iff $\sum_{n=1}^\infty x_n=x_*$ and $\sum_{n=1}^\infty y_n=y_*$. This allows you to solve the problem if you are able to show that the series $\sum_{n=1}^\infty \cos(2\pi n/3) n^{-1/3}$ and $\sum_{n=1}^\infty \sin(2\pi n/3)n^{-1/3}$ converge. Both series are convergent by the Dirichlet's test. For example regarding the first series you have $$\left|\sum_{n=1}^N \cos\left(\frac{2\pi n}{3}\right)\right| = \left|\frac{\cos\left(\frac{2\pi(N+1)}{3}\right) - \cos\left(\frac{2\pi}{3}\right)}{1 - \cos\left(\frac{2\pi}{3}\right)}\right| \leq \frac{1}{\left|\sin\left(\frac{\pi}{3}\right)\right|} = \frac{2}{\sqrt{3}}.$$