The sum of the series
`((a-b)/(a))+1/2((a-b)/(x))^(2)+1/3((a-b)/(a))^(3)+….infty` is

Options

A. `log _(e) ab`
B. `log_(e)((b)/(a))`
C. `log_(e)((a)/(b))`
D. `log_(e) a^(b)`

Answer

To find the sum of the series

\[ S = \frac{(a-b)}{a} + \frac{1}{2} \left(\frac{(a-b)}{a}\right)^2 + \frac{1}{3} \left(\frac{(a-b)}{a}\right)^3 + \ldots \]

we can recognize that this series resembles the Taylor series expansion for the logarithm function.

### Step-by-step Solution:

1. **Identify the Series Structure**:
The series can be rewritten as:
\[ S = \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{(a-b)}{a}\right)^n \]
This is the series for \(-\log(1-x)\) where \(x = \frac{(a-b)}{a}\).

2. **Use the Logarithmic Series**:
Recall that the series for \(-\log(1-x)\) is:
\[ -\log(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n} \]
Therefore, we can substitute \(x\) with \(\frac{(a-b)}{a}\):
\[ S = -\log\left(1 - \frac{(a-b)}{a}\right) \]

3. **Simplify the Argument of the Logarithm**:
Simplifying the expression inside the logarithm:
\[ 1 - \frac{(a-b)}{a} = 1 - \left(1 - \frac{b}{a}\right) = \frac{b}{a} \]

4. **Final Expression**:
Thus, we have:
\[ S = -\log\left(\frac{b}{a}\right) \]
Using the property of logarithms, this can be rewritten as:
\[ S = \log\left(\frac{a}{b}\right) \]

### Conclusion:
The sum of the series is:
\[ S = \log\left(\frac{a}{b}\right) \]