Radius of Convergence of Power Series over Factorial/Complex Case
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Theorem
Let $\xi \in \C$ be a complex number.
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!}$.
Then $\map f z$ converges absolutely for all $z \in \C$.
That is, the radius of convergence of the power series $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ is infinite.
Proof
This is a power series in the form $\ds \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$.
Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that:
\(\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {\frac 1 {\paren {n + 1}!} } {\frac 1 {n!} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {n!} {\paren {n + 1}!} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \cmod {\dfrac 1 {n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Sequence of Powers of Reciprocals is Null Sequence |
Hence the result.
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.4$. Power Series: Example $\text {(ii)}$