# Convergence of a Power Series with Recurring Coefficients to a Rational Function

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:

$$$$\begin{split} \sum_{n=0}^{\infty}a_nz^n & = a_0(1+z^k+z^{2k}+z^{3k}+...) \\ & +a_1(z+z^{k+1}+z^{2k+1}+z^{3k+1}...) \\ & +a_2(z^2+z^{k+2}+z^{2k+2}+z^{3k+2}+...) \\& . \\& . \\& . \\& +a_{k-1}(z^{k-1}+z^{2k-1}+z^{3k-1}+z^{4k-1}+...) \\ & = a_0(1+z^k+z^{2k}+z^{3k}+...) \\ & + a_1z(1+z^k+z^{2k}+z^{3k}+...) \\ & + a_2z^2(1+z^k+z^{2k}+z^{3k}+...) \\& . \\& . \\& . \\ & + a_{k-1}z^{k-1}(1+z^k+z^{2k}+z^{3k}+...) \end{split}$$$$

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:

$$$$\begin{split} \sum_{n=0}^{\infty}a_nz^n & =a_0\left(\sum_{n=0}^{\infty}(z^k)^n\right)+a_1z\left(\sum_{n=0}^{\infty}(z^k)^n\right)+a_2z^2\left(\sum_{n=0}^{\infty}(z^k)^n\right)+...+a_{k-1}z^{k-1}\left(\sum_{n=0}^{\infty}(z^k)^n \right) \\ & = \left(a_0+a_1z+a_2z^2+...+a_{k-1}z^{k-1}\right)\left(\sum_{n=0}^{\infty}(z^k)^n\right) \end{split}$$$$

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 $$+...$$

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

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