To resolve the problem, it is necessary to recognize that both \( A \) and \( B \) represent geometric series. The given series can be expressed in their standard geometric forms:
\[ A = 1 + r^a + r^{2a} + r^{3a} + \dots \]
This is an infinite geometric series with a first term \( T_1 = 1 \) and a common ratio of \( r^a \). The formula for the sum of an infinite geometric series is:
\[ S = \frac{T_1}{1-r} \]
Applying this formula to series \( A \), we get:
\[ A = \frac{1}{1-r^a} \]
(This assumes \( |r^a|<1 \)).
Similarly, series \( B = 1 + r^b + r^{2b} + r^{3b} + \dots \) can be converted to its standard form:
\[ B = \frac{1}{1-r^b} \]
(This assumes \( |r^b|<1 \)).
Next, we equate \( \frac{a}{b} \) to an expression that incorporates \( A \) and \( B \). Utilizing the following relationship1:
\[ \frac{1-r^a}{1-r^b} = \frac{B(A-1)}{A(B-1)} \]
Taking the logarithm of both sides simplifies the expression to:
\[ \frac{a}{b} = \log_\frac{b-1}{b} \left(\frac{A-1}{A}\right) \]
Consequently, the definitive answer is \( \log_\frac{b-1}{b} \left(\frac{A-1}{A}\right) \). This solution is achieved by articulating the ratio \( \frac{a}{b} \) in terms of the series sums \( A \) and \( B \), employing a suitable logarithmic form to connect these expressions.