Sine Function is Absolutely Convergent

Theorem

The real sine function $\sin: \R \to \R$ is absolutely convergent.

Complex Case

The complex sine function $\sin: \C \to \C$ is absolutely convergent.

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.

$\blacksquare$

Also see

• Cosine Function is Absolutely Convergent

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.2$