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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:

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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 $+...$

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    $\begingroup$ That is correct. $\sum_{n=0}^{\infty}a_nz^n$ îs absolutely convergent for $|z| < 1$, therefore the terms can be arbitrarily rearranged. $\endgroup$
    – Martin R
    Commented Oct 9, 2021 at 6:53
  • $\begingroup$ Thanks, @MartinR! $\endgroup$
    – user689775
    Commented Oct 9, 2021 at 7:45

1 Answer 1

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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 $$ (1-z^k) \sum_{n=0}^{\infty}a_nz^n = \sum_{n=0}^{\infty}a_nz^n - \sum_{n=0}^{\infty}a_nz^{n+k} \\ = \sum_{n=0}^{k-1}a_nz^n + \sum_{n=0}^{\infty}\underbrace{(a_{n+k} - a_n)}_{= 0}z^{n+k} = \sum_{n=0}^{k-1}a_nz^n \, . $$

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