Sums of the series $1 + (x^2) / 3! +( x^4) / 5! +\cdots$ How can I compute sum of the series ; $$1 + \frac{x^2}{3!}+\frac{x^4}{5!}+\frac{x^6}{7!}+\frac{x^8}{9!}+\cdots$$
I tried to divide it to two pieces such that $$f(x) = 1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\cdots = \frac{e^x-x}{x}$$ and that $$g(x) = \frac{x}{2!}+\frac{x^3}{4!}+\frac{x^5}{6!}+\frac{x^7}{8!}+\cdots $$
so that answer is equal to $f(x)-g(x)$ but I couldn't proceed since I cannot find any expression for $g(x)$

$$ f(z)= z+\frac{z^3}{3!}+\frac{z^5}{5!}+\ldots $$
is an odd function with $f'(0)=1$ that satisfies the DE $f''(z)=f(z)$, hence $f(z)=\sinh z$ and:
$$ 1+\frac{z^2}{3!}+\frac{z^4}{5!}+\ldots = \frac{\sinh z}{z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2 \pi^2}\right).$$
Notice that by comparing the coefficient of $z^2$ in the RHS and in the LHS we get:
$$\frac{1}{\pi^2}\sum_{n\geq 1}\frac{1}{n^2}=\frac{1}{3!},$$
from which $\zeta(2)=\frac{\pi^2}{6}$ easily follows, for instance.
