Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \frac{1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+n(n+1)}{n^{3}}\) is equal to (A) 1 (B) \(-1\) (C) \(\frac{1}{3}\) (D) None of these

Step-by-Step Solution

Answer

The limit evaluates to \(\frac{1}{3}\). Option (C) is correct.

Step 1: Analyze the Series

The numerator of the given expression is a series where each term is in the form \(n(n+1)\). Thus, the series is \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1)\). We need to express this in a convenient form to evaluate the limit as \(n \to \infty\).

Step 2: Expand and Simplify Terms

Let’s rewrite each term of the series: \(n(n+1) = n^2 + n\). Therefore, the series becomes \(\sum_{k=1}^{n} (k^2 + k)\). This expands to \(\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k\).

Step 3: Use Formula for Summations

Recall the formulas for the sums: \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\). Substitute these into our expression, we have \(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\).

Step 4: Combine and Simplify the Series

Factor out \(\frac{n(n+1)}{6}\):\[\frac{n(n+1)}{6} (2n+1 + 3) = \frac{n(n+1)(2n+4)}{6} = \frac{2n^3 + 6n^2 + 4n}{6}.\] Simplifying gives \(\frac{n^3 + 3n^2 + 2n}{3}\).

Step 5: Evaluate the Limit

Insert this expression into the original limit: \(\lim_{n \to \infty} \frac{n^3 + 3n^2 + 2n}{3n^3}\). Simplify by dividing every term in the numerator by \(n^3\): \(\lim_{n \to \infty} \left( \frac{1}{3} + \frac{3}{3n} + \frac{2}{3n^2} \right)\). As \(n \to \infty, \frac{3}{3n} \to 0\) and \(\frac{2}{3n^2} \to 0\).

Step 6: Conclusion

The only surviving term as \(n \to \infty\) is \(\frac{1}{3}\). Thus, the limit is \(\frac{1}{3}\).

Key Concepts

Series Summation

Series summation involves finding the total value of a series of numbers, particularly useful for understanding the behavior of sequences and their limits. In our problem, the numerator is a series that starts as \(1 \cdot 2, 2 \cdot 3,\) and goes up to \(n(n+1)\). Each term can be expressed as \(n^2 + n\), giving us a new series \(\sum_{k=1}^{n} (k^2 + k)\). This breaks down further into two distinct sums: \(\sum_{k=1}^{n} k^2\) and \(\sum_{k=1}^{n} k\).

To solve the problem, we need to convert these sums into known formula forms. For \(\sum_{k=1}^{n} k^2\), this is \(\frac{n(n+1)(2n+1)}{6}\). For \(\sum_{k=1}^{n} k\), it's \(\frac{n(n+1)}{2}\). These formulae are powerful as they transform complex series into simple expressions, enabling easy manipulation and computation.

Limit Evaluation

Limit evaluation is a technique used to determine the behavior of a function as the input numbers approach a particular value. In this exercise, we aim to calculate \(\lim _{n \rightarrow \infty} \frac{n^3 + 3n^2 + 2n}{3n^3}\).

The strategy is to simplify the expression by dividing each term in the numerator by \(n^3\), which corresponds to the term in the denominator. This gives us \(\frac{1}{3} + \frac{3}{3n} + \frac{2}{3n^2}\). Each part of the sum changes as \(n\) becomes indefinitely large.

• \(\frac{1}{3}\) remains constant because it doesn't depend on \(n\).
• Both \(\frac{3}{3n}\) and \(\frac{2}{3n^2}\) approach zero since the denominators grow without limit.

This limit evaluation helps us conclude that the series tends to \(\frac{1}{3}\) as \(n\) approaches infinity.

Asymptotic Behavior

Asymptotic behavior analyzes how a function behaves as an input grows very large. In terms of the limit problem, we observe how the series and its quotient behave as \(n\) moves towards infinity.

The leading term in both numerator and denominator matters most in asymptotic analysis. In \(\frac{n^3 + 3n^2 + 2n}{3n^3}\), \(n^3\) is the leading term. This means, for large \(n\), the other terms become negligible, simplifying our expression.

Understanding asymptotic behavior is crucial for analyzing limits and series. It allows us to recognize which parts of expressions become insignificant over large domains, therefore, simplifying the problem. By focusing on dominant terms, you can deduce the final outcome, assisting in solving complex calculus problems efficiently.