Find all z such that the series (2z+3)ⁿ / (n log (n+1)) converges

by RoRi

March 17, 2016

Find all complex numbers $z$ such that the series

$\sum_{n=1}^{\infty} \frac{(2z+3)^n}{n \log (n+1)}$

converges.

First, we consider

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(2z+3)^{n+1}}{(n+1) \log (n+2)} \right) \left( \frac{n \log (n+1)}{(2z+3)^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{(2z+3)n \log (n+1)}{n \log(n+2) + \log(n+2)}. \end{align*}$

So, by the ratio test, the series converges if $|2z+3|< 1$ and diverges if $|2z+3|>1$ . If $|2z+3| = 1$ with $Z \neq -1$ then the series also converges by Dirichlet’s test. Hence, the series converges for $|2z+3| \leq 1$ with $z \neq -1$ .