Determine the radius of convergence of ∑ zn / (an + bn)

Determine the radius of convergence $r$ of the power series:

$\sum_{n=1}^{\infty} \frac{z^n}{a^n + b^n} \qquad a > 0, \quad b> 0.$

Test for convergence at the boundary points if $r$ is finite.

Without loss of generality, assume $b \geq a$ . Then, let

$c_n = \frac{z^n}{a^n + b^n}.$

Then we have

$\begin{align*} \lim_{n \to \infty} \frac{c_{n+1}}{c_n} &= \lim_{n \to \infty} \left( \frac{z^{n+1}}{a^{n+1} + b^{n+1}} \right) \left( \frac{a^n + b^n}{z^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{z (a^n + b^n)}{a^{n+1} + b^{n+1}} \\[9pt] &= \lim_{n \to \infty} \frac{ z \left( \left( \frac{a}{b} \right)^n + 1 \right)}{b \left( \left( \frac{a}{b} \right)^{n+1} + 1 \right)}. \end{align*}$

But, $\frac{a}{b} \leq 1$ since $b \geq a$ by assumption. Hence,

$\lim_{n \to \infty} \frac{ z \left( \left( \frac{a}{b} \right)^n + 1 \right)}{b \left( \left( \frac{a}{b} \right)^{n+1} + 1 \right)} = \frac{z}{b}.$

Therefore, $|z| < b$ which implies $r = b$ where $b$ is the larger of $a$ and $b$ . (If $a \geq b$ then the limit is equal to $\frac{z}{a}$ and $r = a$ .)