Manipulation of Infinite Series Example

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.