Sine Function is Absolutely Convergent/Complex Case/Proof 1

Theorem

The definition of the complex sine function is:

$\ds \forall z \in \C: \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$

$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} }$

is convergent.

We have

$\ds \ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} }$

$=$

$\ds \sum_{n \mathop = 0}^\infty \frac {\size z^{2 n + 1} } {\paren {2 n + 1}!}$

$\ds $

$\le$

$\ds \sum_{n \mathop = 0}^\infty \paren{ \frac {\size z^{2 n + 1} } {\paren {2 n + 1}!} + \frac {\size z^{2 n } } {\paren {2 n }!} }$

$\ds $

$=$

$\ds \sum_{n \mathop = 0}^\infty \frac {\size z^n} {n!}$

changing indices

$\ds $

$=$

$\ds \exp \size z$

$\blacksquare$