Manipulation of Infinite Series Example I'm struggling to understand the manipulation of an infinite series shown in the text below.

We begin with the series $\sum_{n=0}^{\infty} u^n$ which we know converges since $|u|<1$. We then express each term as an infinite series using the series for $u$ and the product rule for power series to obtain series for terms with higher exponents.
But then I don't understand on the penultimate line the justification for "unpacking" each of these series and then collecting like terms. I know absolutely convergent series can be rearranged and I know that $\sum_{n=0}^{\infty} u^n$ is absolutely convergent, but this, to my understanding, would only allow you to rearrange the terms in the original series $1+u+u^2+...$ So I could understand repositioning the infinite sums giving each $u^n$. But I can't understand breaking up those infinite sums and then freely repositioning the individual terms.
To put it in a different way, I wouldn't think you could just remove the brackets around for example $(z^2/3!-z^4/5!+...)$ on the third to last line and freely move the terms. I would think those brackets would have to remain and you could just freely move the infinite sum contained in the brackets.
What rule/property of infinite series allows the manipulation shown in the penultimate line of the above text?
 A: What they are trying to do is use $\frac{1}{1-x}=\sum_{r=0}^{\infty}x^{r}$.
You see that inside the radius of convergence the series $\displaystyle u=\sum_{r=1}^{\infty}\frac{(-1)^{r-1}z^{2r}}{(2r-1)!}$ behaves like a single variable. And you CAN view
$\frac{1}{1-u}=\sum_{r=0}^{\infty}u^{r}$.
Then they are just viewing $u^r$ as a multinomial expansion and then just figuring out the first few terms of the series. This is fairly common in Taylor Series/Laurent Series manipulation. All of these has to be done inside the Radius of convergence.
Now for your "brackets" point. We CAN again do this.
for example. If we want to write the Maclaurin Expansion of $e^{\sin(z)}$.
We can do the following.
$\displaystyle e^{\sin(z)}=\sum_{r=0}^{\infty}\frac{\sin(z)^{r}}{r!}=1+\frac{\left(z-\frac{z^{3}}{3!}+...\right)}{1!}+\frac{\left(z-\frac{z^{3}}{3!}+...\right)^{2}}{2!}$  and figure out the first few terms. You will see that this becomes fairly important in something like finding the Residue where only the coeff of $\frac{1}{z}$ will be required.
Now if this method seems too out of place for you.
Then you can just Expand $\csc(z)$ around the origin using it's Maclaurin expansion and then divide by $\frac{1}{z^{2}}$. Computing the first few derivatives is not hard. Also the taylor series for $\csc$ is well known and many references can be found by just a simple google search.
Otherwise what you can also try is dividing $1$ by the polynomial $z-\frac{z^{3}}{3!}+...$ . That is also a form of series manipulation. I would refer you to Brown and Churchill's complex variables and applications
from page 222 under the title Multiplication and Division of Power Series as reference.
I hope this helps.
