Sine Function is Absolutely Convergent

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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:

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


For:

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

to be absolutely convergent we want:

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

to be convergent.


But:

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

is just the terms of:

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

for odd $n$.


Thus:

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


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.

$\blacksquare$


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