I am trying to solve the following: for what values of $z\in \mathbb{C} $ the series $\sum_{k=1}^{\infty} \frac{2k+i}{k+2i} z^k $ converges.
I transformed the series to the form $\sum_{k=1}^{\infty} e^{i(\theta_1 - \theta_2)} z^k$. It easy to see the values for which the series is absolutely convergent, but I have a difficulty to show the convergence of the series without using absolute converges(the values for absolute convergence may be subset of all the values for which the series converges).
My intuition says that since $ e^{i(\theta_1 - \theta_2)}$ is a finite(bounded) number than for the necessary condition of convergence to be satisfied we need $|z|\leq1$. But I am not sure how to prove it formally and if it is correct.
Any help will be great. Thank you.
If we let $\alpha = \limsup \left| \frac{2k+i}{k+2i} \right|^{\frac{1}{k}}$, then the radius of convergence for this series is $R = \frac{1}{\alpha}$. We have $$ \left| \frac{2k+i}{k+2i} \right|^{2} = \frac{4k^2 + 1}{k^2 + 4}.$$ You can show that $1 \leq \frac{4k^2 + 1}{k^2 + 4} < 4$ (try showing this yourself), so $1 \leq \left| \frac{2k+i}{k+2i} \right| < 2$, so $\lim 1^{\frac{1}{k}} \leq \alpha \leq \lim 4^{\frac{1}{k}}$. Hence, $\alpha = 1$ and $R = 1$. Therefore, your series converges for $|z| < 1$, diverges for $|z| > 1$. As for numbers on the unit circle, you have to use a different method to determine whether the series converges or not.