Complex Power Series/Examples/cos i n over n^2

Example of Complex Power Series

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

$a_n = \dfrac {\cos i n} {n^2} z^n$
$S = \ds \sum_{n \mathop \ge 0} a_n z^n$

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

Proof

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

Thus:

 $\ds R$ $=$ $\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n - 1} } {a_n} }$ Radius of Convergence from Limit of Sequence $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {\cos i \paren {n - 1} } {\paren {n - 1}^2} / \dfrac {\cos i n} {n^2} }$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {\cos i \paren {n - 1} } {\cos i n} \dfrac {n^2} {\paren {n - 1}^2} }$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {\exp \paren {i \paren {i \paren {n - 1} } } + \exp \paren {-i \paren {i \paren {n - 1} } } } {\exp \paren {i \paren {i n} } + \exp \paren {-i \paren {i n} } } \dfrac 1 {\paren {1 - \frac 1 n}^2} }$ Cosine Exponential Formulation and simplifying $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {\exp \paren {-\paren {n - 1} } + \exp \paren {n - 1} } {\exp \paren {-n} + \exp \paren n} } \cmod {\dfrac 1 {\paren {1 - \frac 1 n}^2} }$ $\ds$ $=$ $\ds \dfrac {\exp \paren {n - 1} } {\exp \paren n}$ as $\sequence {\dfrac 1 n}$ is a basic null sequence and $e^{-n} \to 0$ $\ds$ $=$ $\ds \dfrac {\exp \paren n \exp \paren {-1} } {\exp \paren n}$ $\ds$ $=$ $\ds \frac 1 e$

$\blacksquare$