Faster way to find Taylor series

Let $g(w) = \sum_{n=0}^{\infty} a_n w^n$.

Then $(w^2+4w+5) \; g(w) = 1$ implies $$\begin{align} 5 a_0 &= 1 \\ 4 a_0 + 5 a_1 &= 0 \\ a_0 + 4 a_1 + 5 a_2 &= 0 \\ a_1 + 4 a_2 + 5 a_3 &= 0 \\ \text{etc.} \end{align}$$

which you can then solve for the $a_n$'s in a stepwise fashion.

Here is one way to find the first few terms for $z_0=2$, using your idea of letting $w=z-2$:

$\displaystyle\frac{1}{1+z^2}=\frac{1}{1+(w+2)^2}=\frac{1}{5+w^2+4w}=\frac{\frac{1}{5}}{1-(-\frac{w^2+4w}{5})}$

$\displaystyle\frac{1}{5}\left(1-\frac{w^2+4w}{5}+\frac{(w^2+4w)^2}{5^2}-\frac{(w^2+4w)^3}{5^3}+\cdots\right)$

$\displaystyle=\frac{1}{5}\left(1-\frac{w^2+4w}{5}+\frac{w^2(w+4)^2}{25}-\frac{w^3(w+4)^3}{125}+\cdots\right)$

$\displaystyle=\frac{1}{5}-\frac{4}{25}w+\frac{11}{125}w^2-\frac{24}{625}w^3+\cdots$

For this particular problem, try a different substitution: $x=z^2$.

Then $$ \frac1{1+x} = \sum (-1)^nx^n$$ so $$ \frac1{1+z^2} = \sum (-1)^nz^{2n}$$

The probelm of finding a closed form is not always easy. If you can find a closed form for the coefficient of $z^k$ in $$ \frac{1}{(1-z)(1-z^2)(1-z^3)(1-z^4)\cdots} $$ tell me about it so I can steal your result, publish it, and become famous. (LOL - this will be a closed form for the partition number of $k$)