If `xgt0` then `(x-1)/(x+1)+(1)/(2)(x^(2)-1)/((x+1)^(2))+(1)/(3)(x^(3)-1)/((x+1)^(3))+....=`

Options

A. `log_(e )x `
B. `log_(e )(1//x)`
C. `log_(e )(1+x)`
D. `log_(e )(1-x)`

Answer

To solve the given series:

\[ S = \frac{x-1}{x+1} + \frac{1}{2} \cdot \frac{x^2 - 1}{(x+1)^2} + \frac{1}{3} \cdot \frac{x^3 - 1}{(x+1)^3} + \ldots \]

we can rewrite each term in the series.

### Step 1: Rewrite the terms
The first term can be rewritten as:
\[ \frac{x-1}{x+1} = \frac{x}{x+1} - \frac{1}{x+1} \]

The second term:
\[ \frac{1}{2} \cdot \frac{x^2 - 1}{(x+1)^2} = \frac{1}{2} \left( \frac{x^2}{(x+1)^2} - \frac{1}{(x+1)^2} \right) \]

The third term:
\[ \frac{1}{3} \cdot \frac{x^3 - 1}{(x+1)^3} = \frac{1}{3} \left( \frac{x^3}{(x+1)^3} - \frac{1}{(x+1)^3} \right) \]

Thus, we can express the series \( S \) as:
\[ S = \left( \frac{x}{x+1} + \frac{x^2}{2(x+1)^2} + \frac{x^3}{3(x+1)^3} + \ldots \right) - \left( \frac{1}{x+1} + \frac{1}{2(x+1)^2} + \frac{1}{3(x+1)^3} + \ldots \right) \]

### Step 2: Identify the series
The first part of the series can be recognized as:
\[ \sum_{n=1}^{\infty} \frac{x^n}{n(x+1)^n} \]
This is the series representation for \(-\log(1 - t)\) where \(t = \frac{x}{x+1}\).

Thus,
\[ \sum_{n=1}^{\infty} \frac{x^n}{n(x+1)^n} = -\log\left(1 - \frac{x}{x+1}\right) = -\log\left(\frac{1}{x+1}\right) = \log(x+1) \]

### Step 3: Evaluate the second part of the series
The second part is:
\[ \sum_{n=1}^{\infty} \frac{1}{n(x+1)^n} = -\log\left(1 - \frac{1}{x+1}\right) = -\log\left(\frac{x}{x+1}\right) = \log(x+1) - \log(x) \]

### Step 4: Combine the results
Now, combining both parts:
\[ S = \log(x+1) - \left(\log(x+1) - \log(x)\right) \]
This simplifies to:
\[ S = \log(x) \]

### Conclusion
Thus, the final result is:
\[ S = \log_e x \]