Proof of $\frac{1}{(1-r)^2}$ I was given that $S=1+2r+3r^2+\cdots = \frac{1}{(1-r)^2}$ was an identity from my students, and I tried to prove directly using geometric series.

I got stuck and looked around online only to be told that taking the derivative of the series that converges to $\frac{1}{1-r}$.

I liked that proof because it was simple and easy, but unfortunately my student is in Alg II so I cannot explain her with derivatives.

My idea was to expand $S$ as $$(1+r+r^2+r^3+\cdots) +(r+r^2+r^3+\cdots)+(r^2+r^3+\cdots)+\cdots$$

but I get confused whether I should rewrite this as $$(1+r+r^2+r^3+\cdots) +r(1+r+r^2+\cdots)+r^2(1+r+\cdots)+\cdots$$

or

$$(1+r+r^2+r^3+\cdots) +((-1)+1+r+r^2+\cdots)+((-1-r)+1+r+\cdots)+\cdots$$

I was able to get the ID using the first way, but I am not sure if the step I took was valid due to some stuff I vaguely remember about manipulating an infinite series like that was not always allowed.

And on the second way I simply did not know where to go from there.

Can I have some help ?

It would also be helpful to know the criteria of when I can manipulate the order of how to add the series.
Your first approach $$1(1+r+r^2+\cdots)+r(1+r+r^2+\cdots)+r^2(1+r+r^2+\cdots)+\cdots$$ is the best one for intuition. Pulling out the common factor then gives us $$(1+r+r^2+\cdots)(1+r+r^2+\cdots)=(1+r+r^2+\cdots)^2,$$ at which point we can use the standard geometric series result for $|r|&lt;1$.

This is perfectly fine to do, since we're dealing with absolutely convergent series, which is equivalent to unconditionally convergent (meaning we can do such rearrangements without fear of changing the result).
Notice that 
$$(1+2r+3r^2+4r^3+\dots)\cdot(1-r)=1+(2-1)r+(3-2)r^2+(4-3)r^3+\cdots=\\
=1+r+r^2+r^3+\cdots=\frac{1}{1-r}\,,$$
which immediatly implies the required identity
$$ 1+2r+3r^2+4r^3+\cdots=\frac{1}{(1-r)^2}\,.$$