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:

$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

For:

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

to be absolutely convergent, we want:

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

to be convergent.

But

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

is just the terms of

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

for even $n$.

Thus:

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

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 $x \in \R$.

Also, the sequence of partial sums $\displaystyle \sum_{n \mathop = 0}^k \frac {\left|{x}\right|^{2n}}{\left({2n}\right)!}$ is increasing.

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

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 16.2$