Cosine Function is Absolutely Convergent

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Theorem

Let $\cos$ be the cosine function.


Then:

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



Complex Case

The complex cosine function $\cos: \C \to \C$ is absolutely convergent.


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).

$\blacksquare$


Sources