Complex Power Series/Examples/2n Factorial over n Factorial Squared

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Example of Complex Power Series

Let $\sequence {a_n}$ be the sequence defined as:

$a_n = \dfrac {\paren {2 n}!} {\paren {n!}^2} z^n$


The complex power series:

$S = \displaystyle \sum_{n \mathop \ge 0} a_n z^n$

has a radius of convergence of $\dfrac 1 4$.


Proof

Let $R$ denote the radius of convergence of $S$.

Thus:

\(\displaystyle R\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n - 1} } {a_n} }\) Radius of Convergence from Limit of Sequence
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {\paren {2 \paren {n - 1} }!} {\paren {\paren {n - 1}!}^2} / \dfrac {\paren {2 n}!} {\paren {n!}^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {\paren {2 n - 2}!} {\paren {2 n}!} \dfrac {\paren {n!}^2} {\paren {\paren {n - 1}!}^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {n^2} {\paren {2 n} \paren {2 n - 1} } }\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {n^2} {4 n^2 - 2 n} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac 1 {4 - \frac 2 n} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 4\) as $\sequence {\dfrac 1 n}$ is a basic null sequence

$\blacksquare$


Sources