Followup-question: Does $ 1 + \rho /2 + \rho^2 /3 + 1/4 + \rho /5 + \rho^2 /6 +... $ with $\rho^3=1$ converge?

Question

In the recent question it was asked whether $ 1 + 1/2 - 1 /3 + 1/4 + 1 /5 - 1 /6 +... $ converges and was answered to the negative.

Just out of curiosity I looked numerically at the series, where the coefficients are the powers of a complex unit-root, so for instance $\rho = \exp(2 \pi i /3)$ and the series is $ 1 + \rho /2 + \rho^2 /3 + 1/4 + \rho /5 + \rho^2 /6 +... $ .

Up to 1000,2000,3000 terms it seems to converge, but is it perhaps easy to answer this in this specific case?

What about the general case where $u_x=\exp(2\pi i x)$ is any number from the complex unit-circle (or at least a m'th complex unit root (of integer or rational order) $\rho_m=\exp(2 \pi i /m$ ) ) and the series looks like $$ s_x = {1 \over u_x}\sum_{k=0}^\infty {u_x^{1+k} \over 1+k}$$ or $$ s_m = {1 \over \rho_m}\sum_{k=0}^\infty {\rho_m^{1+k} \over 1+k}$$ ?

Answer 1

Note that since $\rho^3=1$, we have the series to be $$\dfrac1{\rho} \left(\sum_{k=1}^{\infty} \dfrac{\rho^k}{k}\right) = -\dfrac1{\rho} \ln(1-\rho)$$

Answer 2

The series converges if and only if $1+\rho+\rho^2=0$, i.e. for $\rho=\exp(\pm2\pi\mathrm i/3)$. You can show this by grouping the terms in threes, decomposing the groups into

$$ \frac1{3n-2}+\frac\rho{3n-1}+\frac{\rho^2}{3n}=\frac{1+\rho+\rho^2}{3n-1}+\frac{3n-(3n-2)\rho^2}{(3n-2)(3n-1)3n} $$

and noting that the series for the second term converges and the series for the first term diverges if $1+\rho+\rho^2\ne0$.

Edit in response to the change in the question:

The series $s_x$ converges for any $u_x\ne1$ by Dirichlet's test: The product of a non-increasing sequence $a_n$ of real numbers that converges to zero with a sequence $b_n$ of complex numbers with bounded partial sums converges. In the present case we can use $a_n=\frac1n$ and $b_n=u_x^n$, where $b_n$ is a geometric series with known bounded partial sums if $u_x\ne1$.