Suppose we have a power series of the form $\sum_{n=0}^{\infty}a_nz^n$ with a recurring sequence of coefficients such that $a_{n+k}=a_n$ for all $n$ and some $k\in\mathbb{Z}^+$.
We're supposed to show that the above series converges for $|z|<1$ to a rational function $p(z)/q(z)$ where $q(z)$ has roots all on the unit circle.
I was hoping folks could check my solution below for correctness:
$$$$
First, notice that since $(a_n)$ is recurring, it is bounded in absolute value by some positive real number $M$: $|a_n|\leq M$ for all $n$.
So by the Comparison Test: $$|a_nz^n|=|a_n||z|^n\leq M\cdot|z|^n$$ $\implies\sum_{n=0}^{\infty}a_nz^n$ is absolutely convergent and therefore convergent for $|z|<1$ since $\sum_{n=0}^{\infty}Mz^n$ is abolutely convergent for $|z|<1$ since it is the geometric series.
Now, if $a_{n+k}=a_n$, we can write since $\sum a_nz^n$ is absolutely convergent that:
\begin{equation}
\begin{split}
\sum_{n=0}^{\infty}a_nz^n & = a_0(1+z^k+z^{2k}+z^{3k}+…)
\\ & +a_1(z+z^{k+1}+z^{2k+1}+z^{3k+1}…)
\\ & +a_2(z^2+z^{k+2}+z^{2k+2}+z^{3k+2}+…)
\\& .
\\& .
\\& .
\\& +a_{k-1}(z^{k-1}+z^{2k-1}+z^{3k-1}+z^{4k-1}+…)
\\ & = a_0(1+z^k+z^{2k}+z^{3k}+…)
\\ & + a_1z(1+z^k+z^{2k}+z^{3k}+…)
\\ & + a_2z^2(1+z^k+z^{2k}+z^{3k}+…)
\\& .
\\& .
\\& .
\\ & + a_{k-1}z^{k-1}(1+z^k+z^{2k}+z^{3k}+…)
\end{split}
\end{equation}
So now write $1+z^k+z^{2k}+z^{3k}+…=\sum_{n=0}^{\infty}(z^k)^n$ which converges for $|z|<1$.
Then from above we get:
\begin{equation}
\begin{split}
\sum_{n=0}^{\infty}a_nz^n & =a_0\left(\sum_{n=0}^{\infty}(z^k)^n\right)+a_1z\left(\sum_{n=0}^{\infty}(z^k)^n\right)+a_2z^2\left(\sum_{n=0}^{\infty}(z^k)^n\right)+…+a_{k-1}z^{k-1}\left(\sum_{n=0}^{\infty}(z^k)^n \right)
\\ & = \left(a_0+a_1z+a_2z^2+…+a_{k-1}z^{k-1}\right)\left(\sum_{n=0}^{\infty}(z^k)^n\right)
\end{split}
\end{equation}
But note $\sum_{n=0}^{\infty}(z^k)^n=\frac{1}{1-z^k}$ for some $k\in\mathbb{Z}^+$ since it is a sum of a geometric series with $r=z^k$ and $|z^k|<1$ since we showed earlier that $|z|<1$.
Finally then, we've showed $$\sum_{n=0}^{\infty}a_nz^n=\frac{a_0+a_1z+a_2z^2+…+a_{k-1}z^{k-1}}{1-z^k}=\frac{p(z)}{q(z)}$$
with $q$'s roots clearly being the $k$-th roots of unity all lying on the unit circle $2\pi/k$ radians apart. $\blacksquare$
I guess my only concern with the 'correctness' of the above solution is the decomposition of the power series into sums with $+…$
Best Answer
Your proof is correct because $\sum_{n=0}^{\infty}a_nz^n$ is absolutely convergent for $|z| < 1$, therefore the terms can be arbitrarily rearranged.
An essentially equivalent, but slightly shorter derivation is $$ (1-z^k) \sum_{n=0}^{\infty}a_nz^n = \sum_{n=0}^{\infty}a_nz^n - \sum_{n=0}^{\infty}a_nz^{n+k} \\ = \sum_{n=0}^{k-1}a_nz^n + \sum_{n=0}^{\infty}\underbrace{(a_{n+k} - a_n)}_{= 0}z^{n+k} = \sum_{n=0}^{k-1}a_nz^n \, . $$