It's easy to show with integration by parts that for any n, $\int_0^\infty e^{-x}x^n dx$ is n!
Also for any function that can be represented equivalently as a taylor series around $x=0$, can be written in the form $f(x) = \sum_{n=0}^\infty {f^{(n)}(0)\ x^n\over n!}$.
So for any of these functions, $\int_0^\infty e^{-x}f(x)\ dx = \int_0^\infty e^{-x} \sum_{n=0}^\infty {f^{(n)}(0)\ x^n\over n!}\ dx$, reordering the sum and integral, we get $ \sum_{n=0}^\infty {f^{(n)}(0)\over n!} \int_0^\infty e^{-x}x^n\ dx $, and using the above identity, this simplifies to $\sum_{n=0}^\infty f^{(n)}(0)$
now let $f(x)=e^{-x}$, which follows all requirements to be expanded in a taylor expansion, and so from above $\int_0^\infty e^{-x}*e^{-x}\ dx $ is $\sum_{n=0}^\infty {d^n\over dx}e^{-x}|_0$ or simply $1-1+1-1+1......$
Although this integral can be straight forwardly evaluated as .5
So now this means $1+1-1+1-1+1... = .5$???
So where is the mistake? I know there have been other proofs showing this result but those fault somewhere in the limits taken. So here is the issue that in the limit as n$\rightarrow \infty$ for the n! identity?
Your series is often called Grandi's Series.
You didn't make any mistakes except for trying to apply rules for uniformly convergent series to your series, which yields a known result that has meaning in some contexts but does not convergence in the usual sense.
On the Wikipedia page I linked you will find a number of ways to derive a result of $1/2$, but perhaps the simplest is to take the limit of $\frac{s_n}{n}$ as $n$ goes to infinity, where $s_n$ is the $n$th partial sum of Grandi's Series. In a specific sense, $1/2$ is what the series would be if it were convergent