If S dentes the sum to infinity and `S_n` the sum of n terms of the series `1+1/2+1/4+1/8+….., `such that `S-S_n lt 1/1000` then the least value of n is

Options

A. 8
B. 9
C. 10
D. 11

Explanation

To solve the problem, we need to find the least value of \( n \) such that the difference between the sum to infinity \( S \) and the sum of the first \( n \) terms \( S_n \) is less than \( \frac{1}{1000} \).

### Step-by-Step Solution:

1. **Identify the series**: The series given is \( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \). This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = \frac{1}{2} \).

2. **Find the sum to infinity \( S \)**: The formula for the sum to infinity of a geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

3. **Find the sum of the first \( n \) terms \( S_n \)**: The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Substituting the values:
\[ S_n = 1 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = 1 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\frac{1}{2}} = 2 \left(1 - \left(\frac{1}{2}\right)^n\right) = 2 - \frac{2}{2^n} \]

4. **Set up the inequality**: We need to find \( n \) such that:
\[ S - S_n < \frac{1}{1000} \]
Substituting \( S \) and \( S_n \):
\[ 2 - \left(2 - \frac{2}{2^n}\right) < \frac{1}{1000} \]
Simplifying this gives:
\[ \frac{2}{2^n} < \frac{1}{1000} \]

5. **Solve for \( n \)**: Rearranging the inequality:
\[ 2 < \frac{1}{1000} \cdot 2^n \]
\[ 2000 < 2^n \]
Taking logarithm base 2 on both sides:
\[ n > \log_2(2000) \]

6. **Calculate \( \log_2(2000) \)**: We know that \( 2^{10} = 1024 \) and \( 2^{11} = 2048 \). Therefore:
\[ 10 < \log_2(2000) < 11 \]
This implies that \( n \) must be at least 11.

7. **Conclusion**: The least integer value of \( n \) satisfying the inequality is:
\[ n = 11 \]

### Final Answer:
The least value of \( n \) is \( 11 \).