If `a_(n)=1+(1)/(2)+(1)/(3)+(1)/(4)+(1)/(5)+ . . . .+(1)/(2^(n)-1)`, then

Options

A. `a_(100)lt100`
B. `a_(100)gt100`
C. `a_(200)lt100`
D. none of these

Answer

To solve the problem, we need to analyze the series defined by \( a_n = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2^n - 1} \).

### Step-by-Step Solution:

1. **Understanding the Series**:
The series \( a_n \) is the sum of the reciprocals of the first \( 2^n - 1 \) natural numbers.

2. **Rewriting the Series**:
We can express the series in terms of powers of 2. The series can be rewritten as:
\[ a_n = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2^n - 1} \]

3. **Estimating the Sum**:
To estimate \( a_n \), we can use the fact that the harmonic series \( H_m = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{m} \) is approximately \( \ln(m) + \gamma \) (where \( \gamma \) is the Euler-Mascheroni constant) for large \( m \). Here, \( m = 2^n - 1 \).

4. **Applying the Harmonic Series Approximation**:
Therefore, we can approximate:
\[ a_n \approx \ln(2^n - 1) + \gamma \approx n \ln(2) + \gamma \]
for large \( n \).

5. **Finding Relationships**:
Since \( a_n \) is approximately \( n \ln(2) + \gamma \), we can compare \( a_n \) with \( n \):
\[ a_n < n \quad \text{for sufficiently large } n \]

6. **Evaluating Specific Cases**:
- For \( n = 100 \):
\[ a_{100} < 100 \]
- For \( n = 200 \):
\[ a_{200} < 200 \]

7. **Conclusion**:
Since we have established that \( a_n < n \) for both \( n = 100 \) and \( n = 200 \), we can conclude that the correct option is the one that states \( a_n < n \).

### Final Answer:
The correct option is \( a_{100} < 100 \).