If \( A = 1 + r^a + r^{2a} + r^{3a} + \cdots \infty \) and \( B = 1 + r^b + r^{2b} + r^{3b} + \cdots \infty \), then \( a/b \) is equal to
Show Hint
Always simplify your \( S_\infty \) to find the common ratio \( R \) first. Once you have \( R^a \) and \( R^b \), logarithms are the standard tool to "bring down" the exponents.
Step 1: Understanding the Concept:
The values \( A \) and \( B \) represent the sums of infinite geometric series.
For a geometric series \( a_1 + a_1R + a_1R^2 + \cdots \) to converge, the common ratio \( |R| \) must be less than 1. Step 2: Key Formula or Approach:
The sum of an infinite GP is given by:
\[ S_\infty = \frac{a_1}{1-R} \]
Where \( a_1 \) is the first term and \( R \) is the common ratio. Step 3: Detailed Explanation:
For series \( A \): first term \( a_1 = 1 \), common ratio \( R = r^a \).
\[ A = \frac{1}{1 - r^a} \implies 1 - r^a = \frac{1}{A} \implies r^a = 1 - \frac{1}{A} = \frac{A-1}{A} \]
Taking natural log: \( a \ln(r) = \ln\left(\frac{A-1}{A}\right) \dots (i) \)
For series \( B \): first term \( a_1 = 1 \), common ratio \( R = r^b \).
\[ B = \frac{1}{1 - r^b} \implies 1 - r^b = \frac{1}{B} \implies r^b = 1 - \frac{1}{B} = \frac{B-1}{B} \]
Taking natural log: \( b \ln(r) = \ln\left(\frac{B-1}{B}\right) \dots (ii) \)
Divide equation (i) by equation (ii):
\[ \frac{a \ln(r)}{b \ln(r)} = \frac{\ln\left(\frac{A-1}{A}\right)}{\ln\left(\frac{B-1}{B}\right)} \]
Using the change of base formula for logarithms \( \frac{\ln x}{\ln y} = \log_y(x) \):
\[ \frac{a}{b} = \log_{\frac{B-1}{B}} \left( \frac{A-1}{A} \right) \] Step 4: Final Answer:
The ratio \( a/b \) is equal to \( \log_{\frac{B-1}{B}} \left( \frac{A-1}{A} \right) \).