`(1)/(2) + (3)/(2) . (1)/(4) + (5)/(3).(1)/(8) + (7)/(4).(1)/(16) + ...=` .........

Answer

To find the sum of the series

\[ S = \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{4} + \frac{5}{3} \cdot \frac{1}{8} + \frac{7}{4} \cdot \frac{1}{16} + \ldots \]

we can analyze the pattern in the terms.

### Step 1: Identify the general term
The series can be expressed in terms of \( n \) as follows:
- The \( n \)-th term can be written as:

\[ T_n = \frac{(2n - 1)}{n} \cdot \left(\frac{1}{2}\right)^{n} \]

This is because:
- The numerator follows the pattern \( 1, 3, 5, 7, \ldots \) which can be represented as \( 2n - 1 \).
- The denominator follows the pattern \( 1, 2, 3, 4, \ldots \) which is simply \( n \).
- The factor \( \left(\frac{1}{2}\right)^{n} \) represents the decreasing powers of 2.

### Step 2: Write the sum in summation notation
Now, we can express the sum \( S \) in summation notation:

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

### Step 3: Split the summation
We can split the summation into two parts:

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

This simplifies to:

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

### Step 4: Evaluate the first summation
The first summation is a geometric series:

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

So,

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

### Step 5: Evaluate the second summation
The second summation is related to the logarithm:

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

### Step 6: Combine the results
Now we can combine both results:

\[ S = 2 - \log(2) \]

### Final Answer
Thus, the sum of the series is:

\[ S = 2 - \log(2) \]