For a positive integer n,let a(n) = 1 + frac{ | vedclass

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(D) We analyze the sum $a(n) = \sum_{k=1}^{2^n-1} \frac{1}{k}$.By grouping terms,we can observe that $a(n) = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{

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(D) We analyze the sum $a(n) = \sum_{k=1}^{2^n-1} \frac{1}{k}$.By grouping terms,we can observe that $a(n) = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \dots + \frac{1}{7}) + \dots + (\frac{1}{2^{n-1}} + \dots + \frac{1}{2^n-1})$.Each group of the form $\sum_{k=2^{m-1}}^{2^m-1} \frac{1}{k}$ is greater than $\frac{1}{2}$.Since there are $n$ such groups,$a(n) > \frac{n}{2}$.For $n=200$,$a(200) > \frac{200}{2} = 100$.Also,it is a known property that $a(n) \le n$ for these sums.Thus,$a(100) \le 100$ and $a(200) > 100$ are both true.

For a positive integer $n$,let $a(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^n - 1}$. Then:

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