Sine Function is Absolutely Convergent

Theorem
Let $x \in \R$ be a real number.

Let $\sin x$ be the sine of $x$.

Then:
 * $\sin x$ is absolutely convergent for all $x \in \R$.

Proof
Recall the definition of the sine function:


 * $\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$

For:
 * $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$

to be absolutely convergent we want:
 * $\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } = \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!}$

to be convergent.

But:
 * $\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!}$

is just the terms of:
 * $\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n}{n!}$

for odd $n$.

Thus:
 * $\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!} < \sum_{n \mathop = 0}^\infty \frac {\size x^n} {n!}$

But:
 * $\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n} {n!} = \exp \size x$

from the Taylor Series Expansion for Exponential Function of $\size x$, which converges for all $x \in \R$.

The result follows from the Squeeze Theorem.

Also see

 * Cosine Function is Absolutely Convergent