Convergence of $\sum_{n=0}^{\infty} a_n \overline{z}^n$ based on $\sum_{n=0}^{\infty} a_n z^n$. Let $\sum_{n=0}^{\infty} a_n z^n$ be a convergent series, where $\{a_n\}_{n \in \mathbb{N}} \in \mathbb{R}$ and $z \in \mathbb{C}$. Is $\sum_{n=0}^{\infty} a_n \overline{z}^n$ a convergent series?
This question popped into my mind as I was working on some homework for my analysis class - my gut says that it should be so (moreso I feel it should hold even if $\{a_n\} \in \mathbb{C}$ but that seems much more difficult to prove), but I am struggling to find a proof for this. 
I feel like there's some intuitive or clever way to show that the boundedness of $\sum_{n=0}^{\infty} a_n z^n$ implies the boundedness of $\sum_{n=0}^{\infty} a_n \overline{z}^n$, which is why I restrict the $\{a_n\}$s to reals, as any nice relationship I could find between the bounds of these series is obliterated when their coefficients are complex. 
I am particularly interested in either a proof or counterexample for the case of complex $\{a_n\}$s, but this problem has haunted my sleep for long enough and any proof of either real or complex $\{a_n\}$s would be more than sufficient. Any help would be greatly appreciated! Thanks so much in advance. 
 A: For a counterexample where $a_n \in \mathbb C$, let $a_n = i^n/n$ and $z = i$. Then $a_n z^n = (-1)^n/n$, so $\sum a_n z^n$ is a convergent alternating series. But $a_n \overline{z}^n = 1/n$, so $\sum a_n \overline{z}^n$ is the harmonic series, hence divergent.
A: Note that $\overline{\sum_{n=1}^N a_n z^n} = \sum_{n=1}^N a_n \overline z^n$ since the $a_n$ are reals.
$x\mapsto \overline x$ is a continuous function, hence $\lim_{N\to \infty}\overline{\sum_{n=1}^N a_n z^n}$ exists and is equal to  $ \overline{\sum_{n=1}^\infty a_n z^n}$.
That implies the existence of $\lim_{N\to \infty}\sum_{n=1}^N a_n \overline z^n$, hence convergence of the series.
Furthermore, $$\sum_{n=1}^\infty a_n \overline z^n = \overline{\sum_{n=1}^\infty a_n z^n}$$
A: Yes. Let $S=\sum_{n=0}^\infty a_nz^n$. Then $\overline S=\sum_{n=0}^\infty a_n\overline z^n$ because, if $N\in\mathbb N$,$$\left|\overline S-\sum_{n=0}^Na_n\overline z^n\right|=\left|\overline S-\overline{\sum_{n=0}^Na_nz^n}\right|=\left|\overline{S-\sum_{n=0}^Na_nz^n}\right|=\left|S-\sum_{n=0}^Na_nz^n\right|.$$
A: Let $z=\displaystyle \sum_{n=0}^{\infty} a_nz^n$. By definition we have that for every $\varepsilon >0$, there exists $N \in \mathbb{N}$ such that $$\left \Vert \sum_{n=0}^m a_nz^n-z \right \Vert <\varepsilon, \: \forall m \geq N.$$ Note that $$\left \Vert \sum_{n=0}^m a_n\overline{z^n}-\overline{z} \right \Vert=\left\Vert \sum_{n=0}^m \overline{a_nz^n}-\overline{z} \right\Vert=\left \Vert \overline{\sum_{n=0}^m a_nz^n}-\overline{z} \right\Vert=\left\Vert \overline{\sum_{n=0}^m a_nz^n-z} \right\Vert=\left\Vert \sum_{n=0}^m a_nz^n-z \right\Vert.$$ Hence, for $m \geq N$ we get that $$\left\Vert \sum_{n=0}^m a_n\overline{z_n}-\overline{z}\right\Vert<\varepsilon.$$ So $\sum a_n \overline{z}^n$ converges to $\overline{z}$. 
Note that we used the property $\Vert w \Vert=\Vert \overline{w} \Vert$, for every $w \in \mathbb{C}$, where $\Vert w \Vert:= \sqrt{\Re w+\Im w}$ is the usual norm in the space of complex numbers. 
A: Note that if radius of convergence of the power series is $r$ then $z\in B(0,r)\implies\bar z\in B(0,r)$ too, so if there are problems of convergence this has to be on the border, i.e. on the sphere $S(0,r)$.
This is the case even in the real case, for instance $\sum \dfrac{x^n}n$ converges for $|x|<1$ and for $x=-1$ but not for $x=1$. (of course $\bar 1$ is not $-1$, I just show that problems occur on the border).
In general there is no theorem for the behaviour of the series on $S(0,r)$, this often turns to be a case by case study anyway.
But if your series converges in $\mathbb C$ (i.e. $r$ is infinite), then it doesn't matter whether you consider $z$ or $\bar z$, there is no border.
