Problem 9 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

\(\lim _{n \rightarrow \infty}\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots .+\frac{1}{(2 n+1)(2 n+3)}\right]\) is equal to (a) 0 (b) \(1 / 2\) (c) \(1 / 9\) (d) 2

Step-by-Step Solution

Answer

The limit of the series is \( \frac{1}{2} \).

Step 1: Recognize the Pattern in the Series

The given series \( \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \ldots + \frac{1}{(2n+1)(2n+3)} \) includes terms of the form \( \frac{1}{(2k+1)(2k+3)} \). Recognize that this forms a telescoping series.

Step 2: Decompose the General Term

Each term \( \frac{1}{(2k+1)(2k+3)} \) can be decomposed using partial fraction decomposition. We set \( \frac{1}{(2k+1)(2k+3)} = \frac{A}{2k+1} + \frac{B}{2k+3} \). Solving for \(A\) and \(B\), we find \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \). Thus, \( \frac{1}{(2k+1)(2k+3)} = \frac{1}{2} \left( \frac{1}{2k+1} - \frac{1}{2k+3} \right) \).

Step 3: Express the Series as a Telescoping Series

The series then becomes \( \frac{1}{2} \left( \frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \cdots - \frac{1}{2n+1} + \frac{1}{2n+1} - \frac{1}{2n+3} \right) \). Notice how consecutive terms cancel each other, forming a telescoping series.

Step 4: Simplify and Evaluate the Limit

After cancellation, only the first and last terms remain in effect, yielding \( \frac{1}{2} \left( 1 - \frac{1}{2n+3} \right) \). As \( n \to \infty \), the term \( \frac{1}{2n+3} \to 0 \). Therefore, the limit of the series is \( \frac{1}{2} \).

Key Concepts

Telescoping Series

A telescoping series is a series where most terms cancel out when the series is written out in expanded form. This happens because in these kinds of series, each term has a counterpart that eliminates it, except for a few terms at the beginning or end.
For example, consider the series:

• \( rac{1}{2} \left( \frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \dots \right) \)

In this, every other term cancels with the following, except the first \( \frac{1}{1} \) and the last terms such as \(- \frac{1}{2n+3} \).
This makes telescoping series highly useful in simplifying the series for finding limits as \( n \) approaches infinity. You are typically left with very few surviving terms as \( n \to \infty \), which can then easily be evaluated.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. This is particularly helpful in simplifying expressions for integration or summation.
Let's take our term from the series:

• \( \frac{1}{(2k+1)(2k+3)} \)

We express this as the sum of two fractions (\( \frac{A}{2k+1} + \frac{B}{2k+3} \)) with unknown coefficients \( A \) and \( B \). Solving for these gives specific values for the coefficients, allowing us to rewrite it into a more manageable form:

• \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \), making it: \( \frac{1}{2} \left( \frac{1}{2k+1} - \frac{1}{2k+3} \right) \)

This decomposition is what forms the building block for the cancellation that occurs in telescoping series, linking both concepts closely.

Convergence of Series

The convergence of a series is about whether the sum of its infinite terms approaches a finite value. For a series to be convergent, the terms must diminish to zero as the series progresses.
In the telescoping series, after significant cancellation, you reduced it to a expression like:

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

As \( n \to \infty \), \( \frac{1}{2n+3} \) approaches zero, leaving the limit of the series at \( \frac{1}{2} \).
This means the telescoping series converges to \( \frac{1}{2} \). Achieving convergence is crucial as it confirms that even with an infinite number of terms, the total remains finite and meaningful.