We know for $x \in \mathbb{R}$ that
$$
e^x = \sum_{n=0}^\infty \frac{x^n}{n!},
$$
but what if we were to split the series into the the series containing the even powers of $x$ and the one containing the odd powers of $x$? I.e.,
$$
\sum_{m=0}^\infty \frac{x^{2m}}{(2m)!} + \sum_{m=0}^\infty \frac{x^{2m+1}}{(2m+1)!}.
$$
Firstly, are we guaranteed that each of the series converges (in which case their sum should be $e^x$, I believe)? Secondly, can we write down what each of the series converge to?
Best Answer
Yes, these series converge, absolutely and everywhere. (This is immediate once you know that the series for $e^x$ converges absolutely for all $x$).
The two series are well-studied, and are known as the hyperbolic cosine and sine, respectively.
As noted in comments they have closed expressions, if we consider $e^x$ itself to be closed:
$$ \cosh x := \sum_{n\ge0,\text{ even}} \frac{x^n}{n!} = \frac{e^x+e^{-x}}2 $$
$$ \sinh x := \sum_{n>0,\text{ odd}} \frac{x^n}{n!} = \frac{e^x-e^{-x}}2 $$
More generally, if $f(x)=\sum_{n=0}^\infty a_nx^n$, then leaving out alternate terms in the series always gives $\frac12(f(x)+f(-x))$ and $\frac12(f(x)-f(-x))$, which is the unique way of writing $f$ as a sum of an even function and an odd function.