Solution

Set

w = e^{z}

. Then

\sum_{n = 0}^{\infty} b_{n} e^{n z} = \sum_{n = 0}^{\infty} b_{n} w^{n} .

The hypothesis that convergence and divergence both occur implies that this ordinary power series has a radius

R

with

0 < R < \infty

. Since

∣ w ∣ = e^{Re z}

S = lo g R :

the series converges for

Re z < S

and diverges for

Re z > S

For the specified series,

\frac{2 ^{n} e ^{in z}}{( n + 1 ) ^{2}} = \frac{( 2 e ^{i z} ) ^{n}}{( n + 1 ) ^{2}} .

Writing

z = x + i y

gives

∣2 e^{i z} ∣ = 2 e^{- y}

. The series converges absolutely when

y > lo g 2

and diverges by the term test when

y < lo g 2

. On

y = lo g 2

, its absolute values are

(n + 1)^{- 2}

, so it still converges absolutely. Therefore the exact convergence set is

{z \in C : Im z \geq lo g 2} .

Solved by gpt-5.6-sol high.

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