The sum of the series `1/2x^2+2/3x^3+3/4x^4+4/5x^5+`... is :

Options

A. `(x)/(1+x)+log(1+x)`
B. `(x)/(1-x)+log(1-x)`
C. `-(x)/(1+x)+log(1+x)`
D. none of these

Answer

To find the sum of the series \( S = \frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \frac{4}{5}x^5 + \ldots \), we can manipulate the terms and use known series expansions.

### Step 1: Rewrite the terms
We can express each term in the series in a different form. Notice that:
\[ \frac{n}{n+1} = 1 - \frac{1}{n+1} \]
Thus, we can rewrite the series as:
\[ S = \left(1 - \frac{1}{2}\right)x^2 + \left(1 - \frac{1}{3}\right)x^3 + \left(1 - \frac{1}{4}\right)x^4 + \left(1 - \frac{1}{5}\right)x^5 + \ldots \]
This can be separated into two series:
\[ S = \left(x^2 + x^3 + x^4 + x^5 + \ldots\right) - \left(\frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4 + \frac{1}{5}x^5 + \ldots\right) \]

### Step 2: Sum the first series
The first series is a geometric series:
\[ x^2 + x^3 + x^4 + x^5 + \ldots = x^2 \left(1 + x + x^2 + \ldots\right) = x^2 \cdot \frac{1}{1-x} \quad \text{(for } |x| < 1\text{)} \]
So, we have:
\[ \text{First series sum} = \frac{x^2}{1-x} \]

### Step 3: Sum the second series
The second series can be expressed using the logarithmic series:
\[ \frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4 + \frac{1}{5}x^5 + \ldots = -\ln(1-x) \quad \text{(for } |x| < 1\text{)} \]
This is derived from the series expansion of \(-\ln(1-x)\).

### Step 4: Combine the results
Now we can combine the results from the two series:
\[ S = \frac{x^2}{1-x} + \ln(1-x) \]

### Final Result
Thus, the sum of the series is:
\[ S = \frac{x^2}{1-x} + \ln(1-x) \]