4D (Analysis I)

Solution

\sum_{n \geq 0} a_{n} z^{n}

R = sup {r \geq 0 : the series converges for every ∣ z ∣ < r} .

∣ z ∣ < R

, choose

w

with

∣ z ∣ < ∣ w ∣ < R

. Since

a_{n} w^{n} \to 0

, those terms are bounded, say

∣ a_{n} w^{n} ∣ \leq M

. Therefore

∣ a_{n} z^{n} ∣ \leq M \frac{z}{w}^{n},

∣ z ∣ > R

, convergence would imply absolute convergence at every smaller modulus by the same argument, contradicting the definition of

R

∣ a_{n} ∣ \leq ∣ b_{n} ∣

, convergence of

\sum b_{n} w^{n}

\sum a_{n} z^{n}

whenever

∣ z ∣ < ∣ w ∣

. Letting

∣ w ∣

approach the radius of the

b

-series shows that $R_{a} \geq R_{b}$ .

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