Show that the following series converge uniformly in the given regions: 1) ∑n=1∞zn,0≤|z|&lt;R,R&lt;1 2) ∑n=1∞e−nz,R≤Rez≤1,R&gt;0

1) ∑n=1∞zn,0≤|z|&lt;R,R&lt;1 un(z)=zn By using m-test (i)|un(z)|=|zn|≤|z|n=Rn (ii) The series ∑n=1{∞}Rn =12+p2+… =p1−R geometric series with |R|&lt;1 ⇒∑n=1∞ is convergent Hence, the series ∑n=1∞zn is convergent 2) ∑n=1∞e−nz,R≤Rez≤1,R&gt;0 un(z)=e−nz (i) |vn(z)|=|e−nz|≤e−nx (ii)∑n=1∞e−nx is geometric series with common ratior |e−x|&lt;1 Thus, the series is convergent

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Jessie Jenkins

Answered question

2022-01-25

Show that the following series converge uniformly in the given regions:
1)

\sum_{n = 1}^{\infty} z^{n}, 0 \leq | z | < R, R < 1

\sum_{n = 1}^{\infty} e^{- n z}, R \leq R e z \leq 1, R > 0

Answer & Explanation

rakije2v

Beginner2022-01-26Added 12 answers

\sum_{n = 1}^{\infty} z^{n}, 0 \leq | z | < R, R < 1

u_{n} (z) = z^{n}

By using m-test
(i)

| u_{n} (z) | = | z^{n} | \leq {| z |}^{n} = R^{n}

(ii) The series

\sum_{n = 1} {\infty} R^{n}

= 12 + p^{2} + \dots

= \frac{p}{1 - R}

geometric series
with

| R | < 1

\Rightarrow \sum_{n = 1}^{\infty}

is convergent
Hence, the series

\sum_{n = 1}^{\infty} z^{n}

is convergent
2)

\sum_{n = 1}^{\infty} e^{- n z}, R \leq R e z \leq 1, R > 0

u_{n} (z) = e^{- n z}

(i)

| v_{n} (z) | = | e^{- n z} | \leq e^{- n x}

(ii)

\sum_{n = 1}^{\infty} e^{- n x}

is geometric series
with common ratior

| e^{- x} | < 1

Thus, the series is convergent

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