Cosine Function is Absolutely Convergent

Theorem
Let $\cos$ be the cosine function.

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

Proof
Recall the definition of the cosine function:


 * $\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

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

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

to be convergent.

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

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

for even $n$.

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

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 $x \in \R$.

Also, the sequence of partial sums:
 * $\ds \sum_{n \mathop = 0}^k \frac {\size x^{2 n} } {\paren {2 n}!}$

is increasing.

The result follows from an application of the Monotone Convergence Theorem (Real Analysis).