Radius of Convergence from Limit of Sequence/Real Case

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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:

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

if this limit exists and is nonzero.

If

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

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


Proof

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:

\(\displaystyle \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} }\) \(<\) \(\displaystyle 1\)
\(\displaystyle \iff \ \ \) \(\displaystyle \size {\frac {a_{n + 1} } {a_n} } \size {x - \xi}\) \(<\) \(\displaystyle 1\)
\(\displaystyle \iff \ \ \) \(\displaystyle \size {\frac {a_{n + 1} } {a_n} }\) \(<\) \(\displaystyle \frac 1 {\size {x - \xi} }\)

The result follows from the definition of radius of convergence.

$\blacksquare$


Also presented as

This result can also be seen presented as:

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

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


Sources