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:
- $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
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
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.2$