Explain why you differentiate for a power series

Ok can someone please explain why this is.

I've a series x^3/3 + x^6/6 +x^9/9 + ……
which can be represented as Sigma {n=1 to infinity} x^3n/n
Question is find a simple formula. the answer gives the following:

first step is to differentiate - f’(x)=x^2 + x^5 +x^8 + …. -- Why?????

Provided |x|<1 then this is a sum of an infinite number of terms

Sum = a/(1-r) = x^2/(1 – x^3)
"a" first term = x^2 ---- Why not use 1/3 as per original equation?
"r" the common ratio ---x^3 - this makes sense is in first equation

Therefore f’(x) = x^2/(1 – x^3)

Integrating yields f(x) = -1/3.log(1-x^3) + c - Why integrate

By differentiating, you get rid of the denominator and end up with the sequence being the sum of a geometric series with common ration x^3. There is no common ratio in the original sequence--the ratios are x^3/6, 6/9 x^3, 9/12 x^3 etc.

Then you can use the usual formula for the sum of a geometric series to change it to a closed formula. Then integrating gets you back to the original formula, +/- a constant.

This technique would only work with some sequences, not with just any. What's special about this one is that differentiating leaves you with a geometric sequence.