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PROBLEM

Suppose that the complex series $\displaystyle \sum_{n=0}^{\infty}{a_n}$ converges. Let $r < 1$ and set $D = \{z \in \mathbb{C} : |z| < r\}$.

Prove that $\displaystyle \sum_{n=0}^{\infty}{{a_n}{z^n}}$ converges absolutely and uniformly in $D$.

MY FUTILE ATTEMPTS

I know that I must somehow use the result(s) in this question: If the complex series $\displaystyle \sum_{n=0}^{\infty}{a_n}$ converges, show that there exists a positive number $A$ such that $|a_n| \leq A$ for all $n$..

Also, the easiest way that I can think of to solve this problem is to use the following theorem:

THEOREM (Weierstrass M-Test or Dominated Convergence Theorem)

Given the series of functions $\ \displaystyle \sum_{n=1}^{\infty}{{f_n}(z)}, z \in E$. Suppose that $\{M_n\}$ is a sequence of positive real numbers such that (i) $|{f_n}(z)| \leq M_n, \forall n \in \mathbb{N}, \forall z \in E$. (ii) $\displaystyle \sum_{n=1}^{\infty}{M_n}$ converges. Then $\displaystyle \sum_{n=1}^{\infty}{{f_n}(z)}$ converges absolutely and uniformly on $E$.

Of course I know that I need to set ${f_n}(z) = {a_n}{z^n}$. Then ${f_n}(z)$ is a power series. What I don't know is the sequence $\{M_n\}$.

QUESTIONS

(1) Is the Weierstrass M-Test indeed the best way to tackle this problem? If so, what should be my $M_n$?

(2) If the answer to the first question in (1) is NO, how can I be able to solve this problem?

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    $\begingroup$ You have all the ingredients. You only need to realize that yo can pick $f_n(z)=a_n z^n$ and $M_n= Ar^n$ for the Weierstrass $M$-test, where $A$ is a bound for $(a_n)$. $\endgroup$ – Sangchul Lee Sep 5 '17 at 15:45
  • $\begingroup$ Geez, thanks a bunch @SangchulLee! I knew it was right before my eyes. Just couldn't figure out the proper $M_n$ to use. =) $\endgroup$ – Archimedes Plutonium Sep 5 '17 at 15:53
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    $\begingroup$ No problem, glad it helped :) $\endgroup$ – Sangchul Lee Sep 5 '17 at 15:59
  • $\begingroup$ Archimedes: just a note that you can answer your own question if you want so this question no longer appears to be unanswered. On the other hand, if @SangchulLee submits his response, you will be able to accept it as the answer. $\endgroup$ – Clayton Sep 5 '17 at 20:11
  • $\begingroup$ @Clayton, okay! Posting an answer in a bit. $\endgroup$ – Archimedes Plutonium Sep 6 '17 at 7:10
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We use the Weierstrass $M$-Test.

As hinted in the comments, we just need to pick ${f_n}(z) = {a_n}{z^n}$ and $M_n = Ar^n$, where $A$ is a bound for $\{a_n\}$ (by the result in If the complex series $\displaystyle \sum_{n=0}^{\infty}{a_n}$ converges, show that there exists a positive number $A$ such that $|a_n| \leq A$ for all $n$.). Note that

(i) $\left|{f_n}(z)\right|=\left|{a_n}{z^n}\right|=|a_n||z^n| \leq A{r^n} = M_n, \forall n \in \mathbb{N}, \forall z \in D$.

(ii) $\sum_{n=0}^{\infty}{M_n}$ converges by the Ratio Test.

Therefore, $\sum_{n=0}^{\infty}{{a_n}{z^n}}$ converges absolutely and uniformly in $D$.

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