Solution

Suppose the series converges at

w \neq = 0

. Then its terms

a_{n} w^{n}

are bounded, say by

M

. Whenever

∣ z ∣ < ∣ w ∣

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

so comparison with a geometric series proves absolute convergence at

z

Let

R = sup {∣ w ∣ : \sum_{n = 0}^{\infty} a_{n} w^{n} converges},

with the natural values

0

\infty

allowed. If

∣ z ∣ < R

, choose a convergent point

w

with

∣ z ∣ < ∣ w ∣ \leq R

; the preceding argument gives convergence at

z

. If

∣ z ∣ > R

, convergence at

z

would contradict the definition of

R

. Hence the series converges for

∣ z ∣ < R

and diverges for

∣ z ∣ > R

. No universal assertion is possible on

∣ z ∣ = R

Solved by gpt-5.6-sol high.

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