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. 

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.
