Radius of Convergence of Power Series over Factorial/Complex Case

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Theorem

Let $\xi \in \C$ be a complex number.

Let $\displaystyle \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!}$.


Then $\map f z$ converges absolutely for all $z \in \C$.

That is, the radius of convergence of the power series $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ is infinite.


Proof

This is a power series in the form $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$.

Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that:

\(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {\frac 1 {\paren {n + 1}!} } {\frac 1 {n!} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {n!} {\paren {n + 1}!} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac 1 {n + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle 0\) Sequence of Powers of Reciprocals is Null Sequence

Hence the result.

$\blacksquare$


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