`(1)/(2) + (1)/(4) + (1)/(8) +(1)/(16) + …" to" oo` is

Options

A. `(1)/(2)`
B. 1
C. 2
D. 3

Answer

To find the sum of the infinite series \( S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \), we can recognize that this series is a geometric progression (GP).

### Step-by-Step Solution:

1. **Identify the first term (a)**:
The first term of the series is:
\[ a = \frac{1}{2} \]

2. **Identify the common ratio (r)**:
To find the common ratio \( r \), we can divide the second term by the first term:
\[ r = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2} \]
Alternatively, we can also check by dividing the third term by the second term:
\[ r = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{8} \times \frac{4}{1} = \frac{1}{2} \]

3. **Check the condition for convergence**:
For the sum of an infinite geometric series to converge, the absolute value of the common ratio must be less than 1:
\[ |r| < 1 \]
Here, \( r = \frac{1}{2} \), which satisfies the condition since \( \frac{1}{2} < 1 \).

4. **Use the formula for the sum of an infinite GP**:
The formula for the sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values of \( a \) and \( r \):
\[ S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]

5. **Conclusion**:
Therefore, the sum of the infinite series is:
\[ S = 1 \]