What's $ab$ if we have: $\frac {5030}{5555}=\frac{a+b}{10}+\frac{b}{10^3}+\frac{a}{10^4}+\frac{a+b}{10^5}+\frac{b}{10^7}+\frac{a}{10^8}+\frac{a+b}{10^9}+...$
I have classified the terms of RHS and calculated for example: $$S_1=\frac{1}{10}+\frac{1}{10^5}+\frac{1}{10^9}+...=\frac{10^3}{10^4-1}$$
But suppose we have calculated all of such sums as $S_1,S_2,S_3$ to get:
$$(a+b)S_1+aS_2+bS_3=\frac {5030}{5555}$$
Here I stopped!!
On
As you've found $a+b = 9 \cdot 503$, just substitute $b=9 \cdot 503 - a$ into the final equation and then you should be able to find the values of $b$ and $a$, as well as their product.
On
$$S_1=\frac {10^3}{10^4-1}=\frac {1000}{9999}\\ S_2=\frac 1{1000}S_1; S_3=\frac 1{100}S_1\\ \Rightarrow \frac{1000}{9999}\left[(a+b)+\frac a{1000}+\frac b{100}\right]=\frac {5030}{5555}\\ 1001a+1010b=9054 $$ If $a,b$ are positive integers, then from $$1000(a+b)\;+\;10b\;+\;a=9\;0\;5\;4$$ we can deduce by inspection that one of the solutions is $$a=4, b=5\\ ab=20\;\blacksquare$$
Divide: $$ \frac{5030}{5555} = 0.90549054905490549\dots =\frac{9}{10}+\frac{5}{10^3}+\frac{4}{10^4}+\frac{9}{10^5} +\frac{5}{10^7}+\frac{4}{10^8}+\dots $$ So with $9=a+b, b=5, a=4$, this matches the required form.