Problem 92 - A Complete Resource Book in Mathematics for JEE Main 2017

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

Since the limit of the expression simplifies to 1, the answer is (D) None of these.

Step 1: Understand the Function Inside the Limit

We are given the expression \( n^{-n^2} \left[ (n+1)(n+\frac{1}{2})(n+\frac{1}{2^2})\ldots(n+\frac{1}{2^{n-1}})\right]^n \) and need to evaluate its limit as \( n \to \infty \). This expression consists of a product inside the brackets, raised to the power of \( n \), and then the entirety is multiplied by \( n^{-n^2} \).

Step 2: Simplify the Inner Product

The term inside the brackets is a product of terms: \( (n+1)(n+\frac{1}{2})(n+\frac{1}{2^2})\ldots(n+\frac{1}{2^{n-1}}) \). This can be rewritten using the Gamma function approximation: \( (n+1) \approx n \) and each \( (n+\frac{1}{2^k}) \approx n \) for large \( n \). Therefore, the product approximates to \( n^n \).

Step 3: Combine the Simplified Product with Exponentiation

We have the expression \( [n^n]^n = n^{n^2} \). Substituting this back into the entire expression, we rewrite it as \( n^{-n^2} \times n^{n^2} \).

Step 4: Simplify the Limit Expression

The expression \( n^{-n^2} n^{n^2} \) simplifies to 1 because the powers of \( n \) cancel out. Therefore, the entire expression simplifies to \( 1^n = 1 \).

Step 5: Evaluate the Limit

Since after simplification the expression becomes a constant 1, the limit as \( n \to \infty \) is simply 1.

Key Concepts

Asymptotic Analysis

Asymptotic Analysis is a method in mathematics that helps us understand the behavior of functions as they approach a particular point, often infinity. It's particularly useful in calculus when evaluating limits, as it simplifies complex functions into easier terms as they approach boundaries. In the context of our exercise, when we look at the expression \[n^{-n^2} imes \bigg[ (n+1)(n+\frac{1}{2})(n+\frac{1}{2^2})\ldots\bigg]^nn\bigg]\]as \(n\) approaches infinity, asymptotic analysis lets us focus on terms that truly influence the behavior of the sequence. By approximating, eliminating negligible factors, and concentrating on dominant terms, we're able to determine the relationship between the complexities in the expression. This form of analysis ensures that for very large \(n\), terms like \((n + \frac{1}{2^k}) \approx n\), leading to simplification. Essentially, it justifies why \[(n+1)(n+\frac{1}{2})\ldots \approx n^n\]. Thus, understanding asymptotic behavior is crucial in evaluating limits effectively.

Gamma Function

The Gamma Function, denoted by \(\Gamma(n)\), extends the factorial function to non-integer values. It's defined as: \[n\Gamma(n) = \int_0^{\infty} t^{n-1}e^{-t}\, dtn\]For positive integers, it holds that \(\Gamma(n) = (n-1)!\). In calculus, Gamma Function is handy when dealing with products involving sequences, as it provides a smooth transition from discrete to continuous analysis. In the given exercise, the Gamma approximation \((n+1) \approx n\) and each subsequent term is viewed under the lens of the Gamma function, which lets us predict the behavior of complicated products as \(n\) approaches infinity. This role of weakening the influence of variable adjustments relies on a deeper understanding arrived at through Gamma Relation and approximates the behavior of compounded product trends.

Product Notation

Product Notation, written as \( \prod \), is used to express a series of multiplied terms. Instead of writing each term separately, we encapsulate them into a concise format. This becomes especially useful in the series expressions found in calculus and advanced mathematics. For example, \((n+1)(n+\frac{1}{2})(n+\frac{1}{2^2}) \ldots\) can be succinctly portrayed using product notation: \[\prod_{k=0}^{n-1} \left(n + \frac{1}{2^k}\right)\]. Thus, it streamlines the manipulation of product sequences.In our problem, product notation allows the drastic condensation and visualization of long expressions, leading us to the core simplifications needed. This methodology is effective for identifying patterns and behaviors in products without expanding each term individually, aiding quick evaluations and making it a staple in calculus limit problems and functions.