`2{(m-n)/(m+n)+1/3((m-n)/(m+n))^(3)+1/5((m-n)/(m+n))^(5)`+..}` is equals to

Options

A. `log((m)/(n))`
B. `log((n)/(m))`
C. `log((n)/(m))`
D. none of these

Answer

To solve the given series \( S = 2\left(\frac{m-n}{m+n}\right) + \frac{1}{3}\left(\frac{m-n}{m+n}\right)^3 + \frac{1}{5}\left(\frac{m-n}{m+n}\right)^5 + \ldots \), we can recognize that it resembles the series expansion of the logarithmic function.

### Step-by-Step Solution:

1. **Identify the Series**:
The series can be expressed as:
\[ S = 2x + \frac{1}{3}x^3 + \frac{1}{5}x^5 + \ldots \]
where \( x = \frac{m-n}{m+n} \).

2. **Recognize the Series Pattern**:
The series resembles the Taylor series expansion for \( \log\left(\frac{1+x}{1-x}\right) \):
\[ \log\left(\frac{1+x}{1-x}\right) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots\right) \]

3. **Express \( S \) in Terms of Logarithm**:
From the series expansion, we can write:
\[ S = \log\left(\frac{1 + x}{1 - x}\right) \]

4. **Substitute \( x \)**:
Substitute \( x = \frac{m-n}{m+n} \) into the logarithmic expression:
\[ S = \log\left(\frac{1 + \frac{m-n}{m+n}}{1 - \frac{m-n}{m+n}}\right) \]

5. **Simplify the Expression**:
Calculate the numerator and denominator:
- Numerator:
\[ 1 + \frac{m-n}{m+n} = \frac{(m+n) + (m-n)}{m+n} = \frac{2m}{m+n} \]
- Denominator:
\[ 1 - \frac{m-n}{m+n} = \frac{(m+n) - (m-n)}{m+n} = \frac{2n}{m+n} \]

6. **Combine the Results**:
Now substitute back into the logarithmic expression:
\[ S = \log\left(\frac{\frac{2m}{m+n}}{\frac{2n}{m+n}}\right) = \log\left(\frac{2m}{2n}\right) = \log\left(\frac{m}{n}\right) \]

### Final Result:
Thus, the value of the series is:
\[ S = \log\left(\frac{m}{n}\right) \]