Problem 92 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty} n^{-n^{2}}\left[(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^{2}}\right) \ldots\left(n+\frac{1}{2^{n-1}}\right)\right]^{n}\) (A) \(e\) (B) \(e^{2}\) (C) \(e^{4}\) (D) None of these

Step-by-Step Solution

Answer

(B) \(e^2\)

Step 1: Rewrite the expression

We need to find \(L = \lim_{n \to \infty} n^{-n^2} \left[\prod_{k=0}^{n-1}\left(n + \frac{1}{2^k}\right)\right]^n\).
Note: the terms inside the product are \((n+1), (n+\frac{1}{2}), (n+\frac{1}{4}), \ldots, (n+\frac{1}{2^{n-1}})\).

Step 2: Take the logarithm

Let \(\ln L = \lim_{n \to \infty} \left[-n^2 \ln n + n \sum_{k=0}^{n-1} \ln\left(n + \frac{1}{2^k}\right)\right]\)
\(= \lim_{n \to \infty} n \sum_{k=0}^{n-1} \ln\left(1 + \frac{1}{n \cdot 2^k}\right)\)

Step 3: Approximate using ln(1+x) ≈ x for small x

For large \(n\), \(\frac{1}{n \cdot 2^k}\) is small, so \(\ln\left(1 + \frac{1}{n \cdot 2^k}\right) \approx \frac{1}{n \cdot 2^k}\).
Therefore: \(\ln L \approx \lim_{n \to \infty} n \sum_{k=0}^{n-1} \frac{1}{n \cdot 2^k} = \lim_{n \to \infty} \sum_{k=0}^{n-1} \frac{1}{2^k} = \sum_{k=0}^{\infty} \frac{1}{2^k} = 2\)

Step 4: Compute the limit

Since \(\ln L = 2\), we get \(L = e^2\).
However, let us verify more carefully. The answer choices are (A) \(e\), (B) \(e^2\), (C) \(e^4\), (D) None of these.
Since \(L = e^2\), the answer is (B) \(e^2\).

Key Concepts

Infinite Products

Infinite products are mathematical expressions resembling an infinite sequence of multiplications. In the provided problem, the expression inside the brackets is an example of such a product. The expression \[(n+1)(n+\frac{1}{2})(n+\frac{1}{2^2})\ldots(n+\frac{1}{2^{n-1}})\]is a finite product, which approaches an infinite product as \( n \) tends to infinity. Each term is slightly larger than \( n \), being \( n+\frac{1}{2^k} \), which means as \( n \) increases, the factors become increasingly close to \( n \).

Why is this important? The behavior of infinite products is key to simplifying complex expressions and understanding the growth patterns of mathematical functions. In calculus, infinite products can often be approximated or related to familiar forms. As such, transforming the product into a more manageable form is crucial. Here, the approximation \( n^n \) reflects how for large \( n \), the cumulative impact of seemingly minor adjustments can be effectively represented as a power function.

Asymptotic Analysis

Asymptotic analysis involves studying the behavior of functions as the input grows larger. In our problem, asymptotic analysis helps approximate the expression \[(n+1)(n+\frac{1}{2})(n+\frac{1}{2^2})\ldots(n+\frac{1}{2^{n-1}})\]as \( n^n \cdot (1+\frac{1}{n})(1+\frac{1}{2n})\cdots(1+\frac{1}{(n-1)n}) \).

Essentially, this analysis is a strategy to simplify the behavior of complex expressions in terms of known functions. By examining the problem's limit, it allows us to effectively compare the rates of growth of these products against simpler functions, typically polynomials or exponentials. Recognizing these asymptotic behaviors lets us make sense of how different mathematical processes behave as inputs grow, often leading to significant simplification. Such analyses are crucial, as computational or exact solutions are sometimes impractical for very large values.

Exponential Functions

Exponential functions, such as \( e^x \), feature prominently in calculus due to their unique properties and common appearance in growth and decay models. These functions can often be seen in the limits of infinite sequences or products. In the problem, recognizing an expression's tendency to transform into an exponential function is key to understanding its asymptotic behavior.

The limit expression contains an exponential growth pattern, even if initially disguised as a product. As we simplify and analyze the expression, it reflects the form \[e^{\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{k}}\]which is akin to an exponential function. Calculating such limits often involves taking logarithms, leveraging their ability to transform products into sums, simplifying the analysis.

Why does this matter? Exponential functions describe processes where change accelerates, like compound interest or population growth, making them indispensable in applied mathematics and physics. Understanding and identifying them is therefore pivotal for students engaging with complex calculus problems.