Radius of convergence

For a series ${\sum }_n{a}_n{x}^{m_n}$ with distinct increasing integer exponents, define the ordinary coefficients to be $a_n$ at $m_n$ and zero elsewhere. The Cauchy-Hadamard theorem then givesThe exponent in the root is the actual degree $m_n$, not the index $n$. For $a_n=(n!{)}^2$ and $m_n=n^2$, $0\le 2\log (n!)/n^2\le 2\log n/n\to 0$, so $R=1$. Boundary behavior is a separate question.

For a nonzero polynomial $p$, the power series ${\sum }_{n\ge 0}p(n)z^n$ has radius of convergence $1$. If $p$ has degree $d$ and leading coefficient $c\ne 0$, then $p(n)\sim c{n}^d$ and $|p(n+1)/p(n)|\to 1$, so the ratio test applies inside the unit disk. At every point of its boundary, the terms fail to tend to zero. The zero polynomial is the exceptional case with radius of convergence $\infty$.

Suppose real coefficients satisfy $c\le a_n\le C(1+n{)}^m$ for all sufficiently large $n$, where $c,C>0$ and $m\ge 0$. Then the power series $\sum a_n{z}^n$ has radius of convergence one. For $|z|<1$, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For $|z|\ge 1$, the terms have modulus at least $c$ eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

For $\sum a_n{z}^n$,with the usual conventions for zero and infinity.

If $L=\mathrm{limsup}{a}_n^{1/n}$ for positive coefficients, then the coefficients $a_n^2$ have root limsup $L^2$. Coefficients $a_n^{a_n}$ instead have root termsso their radius depends on the growth of $a_n\log a_n$ relative to $n$.

If the ordinary power series ${\sum }_{n\ge 0}{b}_n{w}^n$ has radius $R\in (0,\infty )$, thenconverges for $\mathrm{Re}z<\log R$ and diverges for $\mathrm{Re}z>\log R$. Behavior on the boundary depends on the coefficients.