Problem 19 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty}\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{n(n+1)}\right)\) is equal to (A) 1 (B) \(-1\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

1

Step 1: Recognize the Series

Notice that the given series is composed of terms of the form \( \frac{1}{n(n+1)} \). To simplify, let's rewrite each term.

Step 2: Simplify Each Term

Each term \( \frac{1}{n(n+1)} \) can be rewritten using partial fraction decomposition: \( \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \). Therefore, the series becomes: \( \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \).

Step 3: Identify the Telescoping Series Pattern

Observe that the series is telescoping. Most terms will cancel out because each negative term \(-\frac{1}{k+1} \) cancels with the following positive term \( \frac{1}{k+1} \).

Step 4: Calculate the Sum After Cancellation

After cancellation, the first term \( \frac{1}{1} \) and the last negative term \( -\frac{1}{n+1} \) remain. The series sum is therefore \( 1 - \frac{1}{n+1} \).

Step 5: Evaluate the Limit

Evaluate the limit as \( n \rightarrow \infty \): \( \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1 - 0 = 1 \).

Step 6: Conclusion

Thus, the overall limit of the series as \( n \) approaches infinity is 1.

Key Concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a powerful technique that helps break down complex fractions into simpler components. It's especially useful when dealing with fractions that have polynomial denominators. The core idea is to express a single complex fraction as the sum of simpler, separate fractions.

In the context of the exercise given, the fraction \( \frac{1}{n(n+1)} \) is decomposed into simpler fractions using partial fraction decomposition:

• The expression is rewritten as \( \frac{1}{n} - \frac{1}{n+1} \).
• This transformation is made possible by finding constants, in this case, 1 and -1, that satisfy the identity for each value of \( n \).

This decomposition allows for easier manipulation and calculation of the series sum, particularly when dealing with telescoping patterns.

Telescoping Series

A Telescoping Series is a type of series where most terms cancel out. This cancellation occurs due to the design of each term being subtracted by the following one, making it perfect for simplifying calculations.

For our exercise, after applying partial fraction decomposition, the series transforms into:

• \( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \)

This pattern means that each negative term cancels the corresponding positive term from the next expression.

Due to cancellation, only the first and the last remaining terms contribute to the final sum, significantly simplifying the solution. Understanding this concept helps visualize why telescoping series are efficient to work with in such problems.

Infinity Limit Evaluation

Infinity Limit Evaluation involves finding the behavior of a function or sequence as the input grows infinitely large. It helps determine the eventual outcome or sum of a series or expression when extended to an infinite number of terms.

In this problem, once the series has been simplified to only the first positive term and the last negative term left uncancelled, it appears as \( 1 - \frac{1}{n+1} \). As \( n \) approaches infinity, the second term \( \frac{1}{n+1} \) approaches zero.

• The result is that the series approaches an overall limit \( \lim_{n \to \infty} \left(1 - \frac{1}{n+1} \right) \).
• This simplifies to \( 1 - 0 \), delivering a limit of 1.

This tells us that the initial infinite series converges to 1. Evaluating limits at infinity is a crucial skill in determining series convergence, particularly in calculus.