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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.)

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