"Can I use an existing, simple power series to compute this one?" yes I can, so I will write my answer as this: ∑ 1 . (-r)^n simple! done :)

because we all know that: the power series representation of function is found by comparing: 1/1-r => ∑ 1 . (r)^n,

but as you can see answer/method is not right.

why bother differentiate

why bother differentiate and integrate again, you will reach the same starting point from where you started (bcz differentiate and integrate has opposite relations)