# Statements > [!NOTE] Theorem > Given a sequence of complex numbers $(a_{n})_{n\ge 0}$, there exists $R \in [0, \infty]$ such that $\sum a_{n}z^{n}$ [[Convergent Real Series|converges]] if $\lvert z \rvert < R$ and diverges if $\lvert z \rvert>R$ (where $\lvert z \rvert$ denotes the complex modulus of $z$). > [!NOTE] Theorem (Radius of convergence II) > Let $\sum_{n=0}^{\infty}a_{n}x^{n}$ be a [[Power Series|univariate real power series about zero]]. One of the following holds: > > (i) The series [[Convergent Real Series|converges]] only if $x = 0$ > > (ii) The series converges for all real numbers $x$. > > (iii) There is a positive real number $R$ with the property that the series converges if $|x|<R$ and diverges if $|x|>R$. > >$R$ is called the *radius of convergence*. In the first case we say that the radius of convergence is $0$ and in the second that it is $\infty$. > >In other words, the set $\left\{ x\in \mathbb{R} : \sum_{n=0}^{\infty} a_{n}t^{n} \right\}$ is an [[Real intervals|interval]] whose midpoint is zero. # Proofs ###### Proof (Analysis 3) The root test asserts that the series of $(a_{n}z^n)_{n \geq 0}$ converges if $\lim \sup \lvert a_{n}z^n \rvert^{1/n}=\lvert z \rvert \lim \sup \lvert a_{n} \rvert^{1/n}$ is less than $1$ and diverges if it is greater than $1$. This implies convergence when $\lvert z \rvert< \frac{1}{\lim\sup \lvert a_{n} \rvert^{1/n}}$ and divergence when $\lvert z \rvert> \frac{1}{\lim\sup \lvert a_{n} \rvert^{1/n}}$ so $R = \frac{1}{\lim\sup \lvert a_{n} \rvert^{1/n}}$. ###### Proof for Real Power Series (Analysis 2) It is clear that the series converges when $x=0.$ Suppose (i) is false then $\sum_{n=0}^{\infty} a_{n}t^{n}<\infty$ for some $t \neq 0.$ Thus by [[Terms of Convergent Series Tend to Zero]], $a_{n}t^{n}\to{0}.$ By [[Convergent Real Sequence is Bounded]], there is some $M>0$ such that $\lvert a_{n} t^{n} \rvert<M$ for all $n\in \mathbb{N}.$ Let $x\in \mathbb{R}$ such that $|x|<|t|.$ Then $|a_{n}x^{n}|=|a_{n}t^{n}| \left\lvert \frac{x}{t} \right\rvert^{n}<M \left\lvert \frac{x}{t} \right\rvert^{n}.$ By [[Convergent Geometric Series]], $\sum M \left\lvert \frac{x}{t} \right\rvert^{n}<\infty$ since $\left\lvert \frac{x}{t} \right\rvert<1.$ So by [[Comparison Test for Series With Non-Negative Terms]], $\sum |a_{n}x^{n}|<\infty.$ By [[Absolutely Convergent Series is Convergent]], $\sum a_{n}x_{n}<\infty$: that is, if $\sum_{n=0}^{\infty}a_{n}t^{n} < \infty$ for some $t\in \mathbb{R},$ $\sum_{n=0}^{\infty} a_{n}x^{n}<\infty$ for all $|x|< |t|.$ Define $R$ as the [[Supremum of Set of Real Numbers|supremum]], $R = \sup \left\{ |t|: \sum a_{n} t^{n} \text{ converges} \right\}.$which exists by [[Real numbers]] since the set is non-empty. Then either $R$ is infinite or finite which gives (ii) and (iii) respectively.