`(x-y)/(x)+1/2((x-y)/(x))^(2)+1/3((x-y)/(x))^(3)+. . . .` is equal to

Options

A. `log_(e)(x-y)`
B. `log_(e)(x+y)`
C. `log_(e)((x)/(y))`
D. `log_(e)xy`

Answer

To solve the given expression

\[ \frac{x-y}{x} + \frac{1}{2} \left(\frac{x-y}{x}\right)^2 + \frac{1}{3} \left(\frac{x-y}{x}\right)^3 + \ldots \]

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

### Step 1: Identify the series
The series can be identified as:
\[ \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{x-y}{x}\right)^n \]
This is the Taylor series expansion for \(-\log(1 - z)\) where \(z = \frac{x-y}{x}\).

### Step 2: Apply the logarithmic series formula
From the logarithmic series, we know that:
\[ -\log(1 - z) = \sum_{n=1}^{\infty} \frac{z^n}{n} \]
Thus, substituting \(z = \frac{x-y}{x}\):
\[ -\log\left(1 - \frac{x-y}{x}\right) = -\log\left(\frac{y}{x}\right) \]

### Step 3: Simplify the logarithmic expression
Now, we simplify:
\[ -\log\left(1 - \frac{x-y}{x}\right) = -\log\left(\frac{y}{x}\right) = \log\left(\frac{x}{y}\right) \]

### Conclusion
Therefore, the value of the original series is:
\[ \log\left(\frac{x}{y}\right) \]

### Final Answer
The expression

\[ \frac{x-y}{x} + \frac{1}{2} \left(\frac{x-y}{x}\right)^2 + \frac{1}{3} \left(\frac{x-y}{x}\right)^3 + \ldots \]

is equal to

\[ \log\left(\frac{x}{y}\right) \]