Radius of Convergence

Definition

Suppose $\sum _{n=0}^{\infty }{a}_n(x-c{)}^n$ is a power series, and let the radius of convergence $R:=\frac{1}{\underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{|a_n}|}$, then $\sum _{n=0}^{\infty }{a}_n(x-c{)}^n$ converges absolutely when $|x-c|<R$, and diverges when $|x-c|>R$ If $a=0$, it is considered $R=\infty$, and if $a=\infty$, it is considered $R=0$

Facts

Let $R$ be the radius of convergence of a series $\sum _{n=0}^{\infty }{a}_n(x-c{)}^n$ The series converges uniformly on any closed interval inside a $(c-R,c+R)$. In other words, If $\forall r,0<r<R$, then the series $\sum _{n=0}^{\infty }{a}_n(x-c{)}^n$ converges uniformly on a $[c-r,c+r]$

If the radius of convergence of a series $\sum _{n=0}^{\infty }{a}_n(x-c{)}^n$ is an $R$, then the radius of convergence of $\sum _{n=1}^{\infty }n{a}_n(x-c{)}^{n-1}$ will also be $R$

If the radius of convergence of a series $\sum _{n=0}^{\infty }{a}_n(x-c{)}^n$ is an $R$, then the radius of convergence of $\sum _{n=0}^{\infty }\frac{a_n}{n+1}(x-c{)}^{n+1}$ will also be $R$