For a positive integer n , let `a(n) = 1+ 1/2 + 1/3 +…+ 1/(2^(n)-1)` : Then

Options

A. `a (100) lt 100`
B. `a (200 ) lt 200`
C. `a(200) gt 100`
D. `a(100) lt 200`

Answer

To solve the problem, we need to analyze the series defined by \( a(n) = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{2^n - 1} \).

### Step-by-step Solution:

1. **Understanding the Series**:
The series \( a(n) \) is the sum of the reciprocals of the first \( 2^n - 1 \) positive integers.

2. **Identifying the Terms**:
We can express \( a(n) \) as:
\[ a(n) = \sum_{k=1}^{2^n - 1} \frac{1}{k} \]

3. **Estimating the Sum**:
The sum of the first \( m \) terms of the harmonic series can be approximated by:
\[ H_m \approx \ln(m) + \gamma \]
where \( \gamma \) is the Euler-Mascheroni constant (approximately 0.577). For our case, we have:
\[ a(n) \approx \ln(2^n - 1) + \gamma \]

4. **Simplifying the Logarithm**:
As \( n \) becomes large, \( 2^n - 1 \) approaches \( 2^n \). Therefore:
\[ \ln(2^n - 1) \approx \ln(2^n) = n \ln(2) \]

5. **Final Approximation**:
Thus, we can approximate \( a(n) \) as:
\[ a(n) \approx n \ln(2) + \gamma \]

6. **Analyzing the Growth**:
Since \( \ln(2) \) is a constant (approximately 0.693), we can see that:
\[ a(n) < n \text{ for sufficiently large } n \]
because \( n \ln(2) < n \) implies \( \ln(2) < 1 \).

7. **Conclusion**:
Therefore, we conclude that \( a(n) < n \) for all positive integers \( n \).

### Analyzing the Options:
- \( a(100) < 100 \) is true.
- \( a(200) < 200 \) is true.
- \( a(200) > 100 \) is false since \( a(200) < 200 \).
- \( a(100) < 200 \) is true since \( a(100) < 100 \).

Thus, the correct options are \( a \), \( b \), and \( d \).