Problem 19 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The limit is 1, so the answer is (A) 1.

Step 1: Understanding the Series

The given limit involves the sum of fractions of the form \( \frac{1}{n(n+1)} \). The series is written as \( \lim_{n \to \infty} S_n \), where \( S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)} \). We need to find the sum of this series as \( n \) approaches infinity.

Step 2: Simplifying the Terms

Each term \( \frac{1}{k(k+1)} \) can be rewritten using partial fraction decomposition. Notice that:\[\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\]This simplification reveals the telescoping nature of the series.

Step 3: Observing the Cancellation

Substitute the simplified form into the series:\[S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\]Here, every intermediary term cancels out, leaving:\[ S_n = 1 - \frac{1}{n+1} \]

Step 4: Evaluating the Limit

Taking the limit as \( n \to \infty \) results in:\[\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1} \right) = 1 - 0 = 1\]Thus, the sum of the series as \( n \) approaches infinity is 1.

Key Concepts

Partial Fraction Decomposition

When dealing with infinite series, particularly those involving products in the denominator, partial fraction decomposition is a powerful tool. It allows us to break down complex fractions into simpler, more manageable pieces. For the series presented in the exercise, each term is of the form \( \frac{1}{k(k+1)} \). Using partial fraction decomposition, we express each term as the difference of two fractions:

• \[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \]

This breakdown is not just a mathematical curiosity; it serves a crucial purpose. By expressing the fraction this way, we lay the groundwork for the series to telescope, which simplifies the process of finding the sum of the series.
In many series, reconstructing terms in terms of partial fractions is the key to spotting patterns that allow for further simplification.

Telescoping Series

The concept of a telescoping series becomes apparent when we substitute the partially decomposed terms back into the series. A telescoping series is one where terms in the sequence cancel out against each other almost entirely.

• For our specific series, substituting the decomposed terms gives us: \[ S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \]

As you can see, each term in the middle cancels with the subsequent term, leaving us only with the first and the last terms of the series. Hence, the name "telescoping," as it reminds one of a retractable telescope where sections collapse into one another.
This cancellation makes the sum so much more straightforward. After collapsing, only a few terms remain, simplifying the calculation significantly.

Convergence of Series

Convergence in series tells us if the infinite sum of terms will result in a finite number. Understanding the convergence is critical as it implies whether the limit exists and what it might be.

• In this exercise, after simplifying the sum using telescoping technique, we noticed that: \[ S_n = 1 - \frac{1}{n+1} \]

To find the limit \( \lim_{n \to \infty} S_n \), we observe that as \( n \) becomes very large, \( \frac{1}{n+1} \) approaches zero. Consequently, the limit of the series becomes:

• \[ \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1 \]

Thus, the series converges to 1.
Recognizing the convergence properties of a series is an essential skill when tackling similar limits or applications in mathematics.