# For which $q\in \mathbb{C}$ is the series $\displaystyle{\sum_n p(n)q^n}$ convergent? [duplicate]

Let $$p:\mathbb{C}\rightarrow \mathbb{C}$$, $$z\mapsto a_kz^k+a_{k-1}z^{k-1}+\ldots +a_1z+a_0$$ be a polynomial with coefficients $$a_j\in \mathbb{C}$$, $$j=0, \ldots ,k$$ and $$a_k\neq 0$$.

For which $$q\in \mathbb{C}$$ is the series $$\displaystyle{\sum_n p(n)q^n}$$ convergent?

We have that \begin{align*}\sum_n p(n)q^n&=\sum_n \left (a_kn^k+a_{k-1}n^{k-1}+\ldots +a_1n+a_0\right )q^n\\ & =a_k \sum_n n^k q^n +a_{k-1} \sum_n n^{k-1} q^n +\ldots +a_1 \sum_n n q^n + a_0 \sum_n q^n \end{align*}

So we have to check the convergence of a series of the form $$\displaystyle{\sum_n n^jq^n}$$.

Let $$a_n=n^jq^n$$.

Then from the ratio test we have that $$\left |\frac{a_{n+1}}{a_n}\right |=\left |\frac{(n+1)^jq^{n+1}}{n^jq^n}\right |=\left |\left (1+\frac{1}{n}\right )^jq\right |$$
To get convergence we want that this is smaller than $$1$$, right?

So we set $$\left |\left (1+\frac{1}{n}\right )^jq\right |<1$$ and solve for $$q$$ ?

For $$n\in\Bbb N$$ we have $$|p(n)|=|a_k|\cdot |n^k|\cdot |1+f(n)|$$ where $$\lim_{n\to\infty}f(n)=0.$$ So $$\lim_{n\to\infty}|p(n)|^{1/n}=1$$ because $$a_k\ne 0.$$ So the Cauchy-Hadamard Radius Formula tells us that the series converges if $$|q|<1$$ and diverges if $$|q|>1.$$
If $$|q|=1$$ and $$k>0$$ then $$|p(n)|\to\infty$$ as $$n\to\infty$$ so the terms $$|p(n)q^n|=|p(n)|$$ do not $$\to 0$$ as $$n\to \infty$$ and the series diverges.
If $$|q|=1$$ and $$k=0$$ then $$p$$ is the non-zero constant $$a_0$$, so $$|p(n)q^n|=|a_0|$$ does not $$\to 0$$ and the series diverges.
The proof of the Cauchy-Hadamard Radius Formula is not hard nor long. It also shows that if a power series converges on a bounded open disc $$D$$ and $$S=\overline S\subset D$$ then the series converges uniformly on $$S$$. There is a nice presentation in Complex Calculus by Ahlfors.