Radius of Convergence from Limit of Sequence/Real Case

Theorem

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

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

Then the radius of convergence $R$ of $S \paren x$ is given by:

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

if this limit exists and is nonzero.

If

$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = 0$

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

From the ratio test, $S \paren x$ is convergent if:

$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} } < 1$

Thus:

$\ds \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} }$

$<$

$\ds 1$

$\ds \leadstoandfrom \ \ $

$\ds \size {\frac {a_{n + 1} } {a_n} } \size {x - \xi}$

$<$

$\ds 1$

$\ds \leadstoandfrom \ \ $

$\ds \size {\frac {a_{n + 1} } {a_n} }$

$<$

$\ds \frac 1 {\size {x - \xi} }$

The result follows from the definition of radius of convergence.

$\blacksquare$

This result can also be seen presented as:

$\ds R = \lim_{n \mathop \to \infty} \size {\frac {a_{n - 1} } {a_n} }$

but in this case the condition under which the radius of convergence is infinite is less conveniently stated.