Radius of Convergence from Limit of Sequence/Real Case

Theorem
Let $\xi \in \R$ be a real number.

Let $\displaystyle S \paren x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.

Then the radius of convergence $R$ of $S \paren x$ is given by:
 * $\displaystyle \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n}$
 * $\displaystyle \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} }$

If either:
 * $\displaystyle \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = 0$
 * $\displaystyle \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n}} = 0$

then the radius of convergence is infinite and therefore the interval of convergence is $\R$.

Proof of First Result

 * $\displaystyle \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n}$:

From the $n$th root test, $S \paren x$ is convergent if $\displaystyle \limsup_{n \mathop \to \infty} \size {a_n \paren {x - \xi}^n}^{1/n} < 1$.

Thus:

The result follows from the definition of radius of convergence.

Proof of Second Result

 * $\displaystyle \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} }$.

From the ratio test, $S \paren x$ is convergent if $\displaystyle \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} } < 1$.

Thus:

The result follows from the definition of radius of convergence.