`(2)/(1).(1)/(3)+(3)/(2).(1)/(9)+(4)/(3).(1)/(27)+(5)/(4).(1)/(81)+…oo=`

Options

A. `log_(e )((3)/(2))`
B. `(1)/(2)+log_(e )((3)/(2))`
C. `(1)/(2)`
D. `(1)/(2)-log_(e )((3)/(2))`

Answer

To solve the series

\[ S = \frac{2}{1} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{1}{9} + \frac{4}{3} \cdot \frac{1}{27} + \frac{5}{4} \cdot \frac{1}{81} + \ldots \]

we can rewrite the terms in a more manageable form.

### Step 1: Rewrite the terms
The series can be expressed as:

\[ S = \left( \frac{2}{1} \cdot \frac{1}{3} \right) + \left( \frac{3}{2} \cdot \frac{1}{9} \right) + \left( \frac{4}{3} \cdot \frac{1}{27} \right) + \left( \frac{5}{4} \cdot \frac{1}{81} \right) + \ldots \]

This can be rewritten as:

\[ S = \sum_{n=1}^{\infty} \frac{n+1}{n} \cdot \frac{1}{3^n} \]

### Step 2: Split the terms
We can separate the series into two parts:

\[ S = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n} \]

### Step 3: Calculate the first series
The first series is a geometric series:

\[ \sum_{n=1}^{\infty} \frac{1}{3^n} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

### Step 4: Calculate the second series
The second series can be recognized as the series for \(-\log(1-x)\):

\[ \sum_{n=1}^{\infty} \frac{x^n}{n} = -\log(1-x) \]

For \(x = \frac{1}{3}\):

\[ \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n} = -\log\left(1 - \frac{1}{3}\right) = -\log\left(\frac{2}{3}\right) \]

### Step 5: Combine the results
Now we combine both parts:

\[ S = \frac{1}{2} - \left(-\log\left(\frac{2}{3}\right)\right) = \frac{1}{2} + \log\left(\frac{3}{2}\right) \]

### Final Answer
Thus, the final answer for the series is:

\[ S = \frac{1}{2} + \log\left(\frac{3}{2}\right) \]