Animated Solution for Mathematics - Sequence and Series: For a positive integer $n$, let $a(n)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{(2^n)-1}$. Then

A Journey Through Bounds

Welcome, fellow traveler, to one of the most beautiful and deceptive landscapes in mathematics: the Harmonic Series. At first glance, the expression

looks like a simple addition problem. But as any JEE aspirant knows, the harmonic series is a wolf in sheep's clothing. It grows, and it grows, and it never stops. To tame this beast, we don't calculate; we bound.

Imagine you are standing before this infinite-looking sum. You cannot sum it directly. So, what do you do? You organize. You create buckets.

Look at the structure: the series ends at $2^n-1$. This is not a coincidence; it is a roadmap. We will group the terms into blocks where the number of terms in each block is a power of two.

We leave the first term, $1$, alone. Then, we take the next two terms: $(\frac{1}{2}+\frac{1}{3})$. Then, the next four terms: $(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})$.

This continues until we have $n-1$ such groups. This is the Grouping Strategy, and it is your most powerful weapon in the JEE arsenal.

Taming the Beast

To find an upper bound, we want to make our sum larger. How? By replacing every term in a group with the largest term in that group.

For the group $(\frac{1}{2}+\frac{1}{3})$, the largest term is $\frac{1}{2}$. So, we replace both with $\frac{1}{2}$. The sum becomes:

For the next group, $(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})$, the largest term is $\frac{1}{4}$. Replacing all four terms with $\frac{1}{4}$ gives us:

Do you see the elegance? Each group, no matter how large, sums to less than $1$. Since we have $n-1$ such groups, plus the initial $1$, we get the beautiful inequality:

This immediately tells us that for $n=100$, $a(100)<100$. Option A is locked in!

The Flip Side

Now, let's flip the logic. To find a lower bound, we replace every term in a group with the smallest term in that group.

For the group $(\frac{1}{2}+\frac{1}{3})$, the smallest term is $\frac{1}{3}$. But to make the math even cleaner, we use the next power of two: $\frac{1}{4}$. So:

For the next group, $(\frac{1}{4}+\cdots +\frac{1}{7})>4\times \frac{1}{8}=\frac{1}{2}$. Each group is now strictly greater than $\frac{1}{2}$.

With $n-1$ groups, our sum becomes:

We have successfully trapped the harmonic sum between two walls. For $n=200$, our lower bound gives us:

Since $100.5>100$, we have proven that $a(200)>100$. Option D is also correct!

Take a moment to appreciate this. We didn't need a calculator or a computer. We used pure, logical grouping to conquer a problem that seems to defy simple arithmetic.

This is the heart of JEE mathematics: finding the hidden structure in the chaos. Keep this 'grouping' technique in your toolkit—it will serve you well in many more challenges to come.