Product of two geometric series

Question

I have used the Product of two power series and find out the below results. But it is to some extend strange for me, could you please confirm the results?

Let $A=\sum_{i=0}^{\infty}(\frac{L}{a})^i$ and $B=\sum_{i=0}^{\infty}(\frac{L}{b})^i$, then $AB$ is a linear function of $A$ and $B$. \begin{align} AB & =\sum_{i=0}^{\infty}(\frac{L}{a})^i \sum_{i=0}^{\infty}(\frac{L}{b})^i \nonumber\\ &= \sum_{i=0}^{\infty}\sum_{k=0}^{i}(\frac{L}{a})^k(\frac{L}{b})^{i-k} \nonumber\\ &= \sum_{i=0}^{\infty}(\frac{L}{b})^i\sum_{k=0}^{i}(\frac{b}{a})^k \nonumber\\ &= \sum_{i=0}^{\infty}(\frac{L}{b})^i(\frac{1-(\frac{b}{a})^{i+1}}{1-(\frac{b}{a})}) \nonumber\\ &= \frac{a}{a-b}\sum_{i=0}^{\infty}(\frac{L}{b})^i+ \frac{b}{b-a}\sum_{i=0}^{\infty}(\frac{L}{a})^i. \qquad \text{ if } a>b \end{align} it is strange! is not it? Note that the above prove does not work for finite series. (test it)

Answer 1

Let us admit that $A$ and $B$ are finite numbers. So, we can write for the geometric series $$A=\sum_{i=0}^{\infty}(\frac{L}{a})^i=\frac{a}{a-L}$$ $$B=\sum_{i=0}^{\infty}(\frac{L}{b})^i=\frac{b}{b-L}$$ As a result $$ AB =\sum_{i=0}^{\infty}(\frac{L}{a})^i \sum_{i=0}^{\infty}(\frac{L}{b})^i=\frac{a b}{(a-L) (b-L)} =\frac{a b}{(a-b) (b-L)}-\frac{a b}{(a-b) (a-L)}$$ that is to say $$ AB =\frac{a \text{B}}{a-b}-\frac{b \text{A}}{a-b}$$ $$$$$$$$