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:

$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$

For:

$\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!}$

to be absolutely convergent we want:

$\displaystyle \sum_{n \mathop = 0}^\infty \left|{\left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!} }\right| = \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2 n + 1} } {\left({2 n + 1}\right)!}$

to be convergent.

But:

$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2 n + 1} } {\left({2 n + 1}\right)!}$

is just the terms of:

$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n}{n!}$

for odd $n$.

Thus:

$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2 n + 1} } {\left({2 n + 1}\right)!} < \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n} {n!}$

But:

$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n} {n!} = \exp \left|{x}\right|$

from the Taylor Series Expansion for Exponential Function of $\left|{x}\right|$, 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): $\S 16.2$