The function is
$$f(x)=e^{-x^2}$$
Now using the power series formulas of $e^x$ and manipulating it to become the given function it show be
$$\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{n!}$$
Now if I want to evaluate the integral of the function and give my answer in sum notation
I would get
$$\sum_{n=0}^\infty\frac{-x^{2n+2}}{(n+1)!}$$
Sorry I keep editing the question, there is a lot of parts to this particular question.
Since the Maclaurin series of $e^x$ is $$ \sum_{k=0}^{\infty}\frac{x^k}{k!} $$ one gets the Maclaurin series of $f(x)=e^{-x^2}$:
$$ \sum_{k=0}^{\infty}\frac{(-1)^k}{k!}x^{2k} $$