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

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 is \(\frac{1}{3}\), option (C).

Step 1: Recognize the Numerator Pattern

Observe that the numerator of the expression represents a series of terms: \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n + 1)\). These terms can be written as \(n(n + 1)\). This series is equivalent to \(\sum_{k=1}^{n} k(k+1)\).

Step 2: Simplify the Summation

Simplify each term in the summation: \(k(k + 1) = k^2 + k\). So, \(\sum_{k=1}^{n} k(k+1) = \sum_{k=1}^{n} (k^2 + k) = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k\).

Step 3: Use Known Summation Formulas

Apply the formulas for the sums of squares and integers. The formula for the sum of squares \(\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}\), and for integers \(\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}\).

Step 4: Substitute Summation Formulas

Substitute these formulas into the expression: \[\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\]

Step 5: Combine and Simplify

The expression becomes:\[\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} = \frac{n(n + 1)[(2n + 1) + 3]}{6} = \frac{n(n + 1)(2n + 4)}{6}\]Simplify further: \[\frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)2(n+2)}{6} = \frac{n(n+1)(n+2)}{3}\]

Step 6: Express in Terms of Non-Dominant Terms

Focus on the dominant power of \(n\) in the terms: \(\frac{n^3 + 3n^2 + 2n}{3}\).

Step 7: Divide by Dominant Term

Now consider the limit:\[\lim_{n \to \infty} \frac{\frac{n^3 + 3n^2 + 2n}{3}}{n^3} = \lim_{n \to \infty} \frac{n^3 + 3n^2 + 2n}{3n^3} = \lim_{n \to \infty} \left(\frac{1}{3} + \frac{3}{3n} + \frac{2}{3n^2}\right) = \frac{1}{3}\].

Step 8: Evaluate the Limit at Infinity

As \(n\) approaches infinity, the terms \(\frac{3}{3n}\) and \(\frac{2}{3n^2}\) approach zero. This leaves \(\frac{1}{3}\) as the final result.

Key Concepts

Summation formulas

Understanding summation formulas is crucial in calculus, especially when dealing with series. In this exercise, we encounter the formulas for the sum of squared integers and the sum of integers. These formulas appear frequently and help transform complex series into manageable expressions.

• For the sum of the first n integers, \( \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \).
• For the sum of squares of the first n integers, \( \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \).

These formulas allow us to replace large sums with simpler polynomial expressions, making it easier to manipulate and solve problems involving limits.

Series simplification

Series simplification is the process of breaking down complex series into simpler parts. In this exercise, we did that by recognizing the numerator pattern and using the applicable summation formulas.

• First, identify repeating patterns: here, \( k(k+1) \) translates to \( k^2 + k \).
• Substitute summation formulas for these patterns: converting series into sums \( \sum_{k=1}^{n} (k^2 + k) = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \).

Series simplification reduces the number of terms and simplifies the expressions involved, making them much easier to evaluate.

Limit evaluation

Limit evaluation involves determining the value that a function approaches as the input approaches a certain point or infinity. In our exercise, we evaluated the limit as \( n \) approaches infinity.

• First, simplify the expression as much as possible.
• Focus on finding the dominant terms to understand what the function behaves like as \( n \) becomes very large.
• Evaluate the limit by simplifying the polynomial: \( \frac{n^3 + 3n^2 + 2n}{3n^3} \).

By carefully breaking down the expression and focusing on the dominant term, we successfully determine that the limit is \( \frac{1}{3} \).

Dominant terms

Dominant terms in calculus refer to the terms in an expression that have the highest power of the variable. When \( n \) is large, these terms impact the behavior of the function most significantly.

• Identify the highest power of \( n \) in both the numerator and the denominator.
• Simplify the expression by comparing dominant terms: here \( \frac{n^3}{n^3} = 1 \).
• The presence of lower power terms like \( \frac{3n^2}{3n^3} \) becomes negligible as \( n \) approaches infinity.

The focus on dominant terms simplifies limit calculations as only the most impactful terms are considered, leaving negligible terms out.

Mathematical proofs

Mathematical proofs are logical arguments that validate a mathematical statement. In our limit evaluation, we logically structured each step:

• Start with a clear statement of the problem: understanding the sequence and what is being summed.
• Apply mathematical rules such as summation formulas to expand and simplify the expression.
• Demonstrate how smaller terms became negligible, leading to the final limit value of \( \frac{1}{3} \).

Each step serves as a piece of the puzzle, contributing to a comprehensive proof of the exercise's solution. This logical structure ensures all aspects are verified and understood.