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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 $q$ ?

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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.

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  • $\begingroup$ a.k.a. the Hadamard Radius Formula. Discovered by Cauchy, discovered again decades later by Hadamard. $\endgroup$ Jan 26 at 6:28
  • $\begingroup$ Thanks a lot!! :-) $\endgroup$
    – Mary Star
    Feb 16 at 21:20

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