Complex Power Series/Examples/3^n-1 over 2^n+1

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Example of Complex Power Series

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

$a_n = \dfrac {3^n - 1} {2^n + 1}$


The complex power series:

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

has a radius of convergence of $1$.


Proof

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

By Radius of Convergence from Limit of Sequence:

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


Thus:

\(\displaystyle a_n\) \(=\) \(\displaystyle \dfrac {3^n - 1} {2^n + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\frac {3^n} {2^n} - \frac 1 {2^n} } {1 + \frac 1 {2^n} }\) multiplying top and bottom by $\dfrac 1 {2^n}$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\frac 3 2 - \frac 1 {2^n} } {1 + \frac 1 {2^n} }\)
\(\displaystyle \) \(\to\) \(\displaystyle \dfrac {3 / 2} 1\) as $n \to \infty$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 3 2\) and so is independent of $n$


Thus:

\(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n - 1} } {a_n} }\) \(=\) \(\displaystyle \cmod {\dfrac {3 / 2} {3 / 2} }\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$



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