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Chapter 2: Problem 9

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

Short Answer

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

Step by step solution

01

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.
02

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, \( \frac{1}{(2k+1)(2k+3)} = \frac{1}{2} \left( \frac{1}{2k+1} - \frac{1}{2k+3} \right) \).
03

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.
04

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 \( \frac{1}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

If \(\lim _{x \rightarrow \infty}\left[\frac{x^{3}+1}{x^{2}+1}-(a x+b)\right]=2\), then (a) \(a=1\) and \(b=1\) (b) \(a=1\) and \(b=-1\) (c) \(a=1\) and \(b=-2\) (d) \(a=1\) and \(b=2\)

If \(f(x)=\left\\{\begin{array}{ll}x, & x<0 \\ 1, & x=0 \\ x^{2}, & x>0\end{array}\right.\), then \(\lim _{x \rightarrow 0} f(x)=\) (a) Is 1 (b) Is zero (c) Does not exist (d) None of these

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