Polynomial Long Division for the sequence of a geometric series?

Question

Why does polynomial long division only work for power series if the leading term of the denominator is smaller than the leading term of the numerator?

For instances I see how long division can work for $$ \frac{1}{1-3z}$$

However, Polynomial long divisions will not work for something such as $$ (\frac{1}{2-z})$$

When trying long division for $$ \frac{1}{2-z}$$, the first couple of terms I got were $$ \frac{1}{2}+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}... $$

Although by factoring $$ \frac{1}{2-z}$$ to $$ \frac{1}{2}(\frac{1}{1-\frac{z}2})$$ I can replace the z in the standard geometric series with $$ \frac{z}{2}$$ and then multiply each term by 1/2, I see that the correct terms are $$ \frac{1}{2} +\frac{z}{4} + \frac{z^2}{8} + \frac{z^3}{16}... $$

Why does polynomial long division not work in cases where the leading term in the denominator is larger then the leading term in the numerator?

Answer 1

Polynomial long division would work for $1/(2-z)$ if you did it right.

        1/2 + z/4 + ...
      -------------
2 - z ) 1
        1 - z/2
        -------
            z/2 
            z/2 - z^2/4
            -----------

etc

Answer 2

You've done the long division for $\frac1{2-z}$ wrongly: $$1-\frac12(2-z)=\frac z2$$ $$\frac z2-\frac z4(2-z)=\frac{z^2}4$$ $$\frac{z^2}4-\frac{z^2}8(2-z)=\frac{z^3}8\cdots$$