Problem 71 - A Complete Resource Book in Mathematics for JEE Main

Question

For a positive integer \(n\), let \(a(n)=\) \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{\left(2^{n}\right)-1}\), Then (A) \(a(100) \leq 100\) (B) \(a(100)>100\) (C) \(a(200) \leq 100\) (D) \(a(200)>100\)

Step-by-Step Solution

Answer

(A) and (D) are correct.

Step 1: Understanding the Function

The function given is \(a(n) = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{(2^n) - 1}\). This is a partial sum of a harmonic series. The important thing to note is that the series starts from \(1\) and goes up to the \((2^n - 1)\)-th term.

Step 2: Approximate Harmonic Series Growth

The complete harmonic series \(H_m = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{m}\) grows approximately as \(\ln(m) + \gamma\), where \(\gamma\) is the Euler-Mascheroni constant, roughly equal to 0.577. For large \(n\), \(a(n) \approx \ln((2^n) - 1) + \gamma\).

Step 3: Apply for \(a(100)\)

Calculate for \(n = 100\):\[a(100) = \ln((2^{100}) - 1) + \gamma \approx 100 \cdot \ln 2 + \gamma \approx 100 \cdot 0.693 + 0.577 = 69.3 + 0.577 \approx 69.877\]Since \(69.877 < 100\), option (A) is correct: \(a(100) \leq 100\).

Step 4: Apply for \(a(200)\)

Calculate for \(n = 200\):\[a(200) = \ln((2^{200}) - 1) + \gamma \approx 200 \cdot \ln 2 + \gamma \approx 200 \cdot 0.693 + 0.577 = 138.6 + 0.577 \approx 139.177\]Since \(139.177 > 100\), option (D) is correct: \(a(200) > 100\).

Key Concepts

Logarithmic Growth

When we talk about logarithmic growth, we are referring to a way in which certain quantities increase. Logarithmic growth occurs when a quantity grows in proportion to the logarithm of another. This type of growth is slower than linear or polynomial growth. It is often applied in scenarios where the quantity expands rapidly at first but slows down over time.
In the context of the harmonic series, as seen with the function \(a(n)\), the partial sum \(1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{m}\) is approximately equal to the logarithm of \(m\), specifically \(\ln(m)\). This relationship shows how the harmonic series represents logarithmic growth. The larger \(m\) becomes, the sum's growth pace diminishes, as dictated by the logarithmic nature.

Euler-Mascheroni Constant

The Euler-Mascheroni constant, denoted by \(\gamma\), is a significant number in mathematics, much like \(\pi\) and \(e\). Its approximate value is 0.577. This constant emerges in calculus, number theory, and Euler's work on integrating the harmonic series.
In connection with the harmonic series, it appears in the approximation formula \(H_m \approx \ln(m) + \gamma\). Here, \(H_m\) is the sum of the first \(m\) terms of the harmonic series. The Euler-Mascheroni constant accounts for the discrepancy between the natural logarithm \(\ln(m)\) and the actual sum of the series. Understanding \(\gamma\) is essential to grasp the asymptotic analysis of these series.

Partial Sum Approximation

Partial sum approximation is a technique used to estimate the sum of a series up to a certain number of terms. Especially for series that converge slowly or increase without bound, like the harmonic series, partial sums are practical. They allow us to estimate the value of the series to a reasonable degree of accuracy.
In the given function \(a(n)\), the sum is calculated only up to \((2^n - 1)\)-th term. To approximate it, we use the relation \(a(n) \approx \ln((2^n) - 1) + \gamma\). This provides a manageable estimate by transforming a potentially complex sum into terms of logarithmic and constant values. These approximations simplify calculations, especially when evaluating series sums for large values of \(n\). By using effective approximations, complex series can be handled more easily, offering insights into the series' behavior as \(n\) grows larger.