Closed form of power series First day on the site and this is an amazing place!
this is my question: 
knowing that: $$1+\frac{x^2}{2}+\frac{x^4}{4}+\frac{x^6}{6}+\frac{x^8}{8}+\cdots $$
is a power series, how can I obtain its closed form?
Thank you.

David's answer is certainly the way to go. I will just expand a little about the result (for $|x|<1$)
$$
S =x + x^3 + x^5 + \cdots = \frac{x}{1-x^2} 
$$
Notice that in such case both series:
$$
S_1 =1 + x + x^2 +\cdots = \frac{1}{1-x}
$$
$$
S_2 = 1 + x^2 + (x^2)^2 + (x^2)^3 + \cdots = \frac{1}{1-x^2}
$$
converge absolutely, so we may write $S = S_1-S_2$ and substract termwise, one gets:
$$
S = \frac{1}{1-x} -\frac{1}{1-x^2} =\frac{x}{1-x^2}
$$ 
