Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a nice formula for the sum $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} $$

Step-by-Step Solution

Answer

The sum is \( 1 - \frac{1}{n+1} \).

Step 1: Identify the General Term

The given sum is \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{n(n+1)} \). We observe that the general term of the sequence is \( \frac{1}{k(k+1)} \) for \( k \) ranging from 1 to \( n \).

Step 2: Decompose the General Term

To simplify \( \frac{1}{k(k+1)} \), we use partial fraction decomposition. We can express it as \( \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \). This formula can be verified by finding a common denominator and observing the cancellation of terms.

Step 3: Re-write the Series Using the Decomposed Form

Substitute the decomposed form of each term back into the series: \( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \).

Step 4: Notice the Telescoping Nature

Recognize that the series is telescoping, meaning that most intermediate terms cancel out. Specifically, positive fractions cancel with negative fractions in successive terms, leaving only the first term of the first fraction and the second term of the last fraction.

Step 5: Calculate the Remaining Terms

After cancellation, only the first term from the first fraction \( \frac{1}{1} \) and the negative term from the last fraction \( -\frac{1}{n+1} \) remain. Thus, the sum simplifies to \( 1 - \frac{1}{n+1} \).

Step 6: Final Formula

The sum \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{n(n+1)} \) can be expressed concisely with the formula: \( 1 - \frac{1}{n+1} \).

Key Concepts

Partial Fraction Decomposition

In the world of calculus and algebra, partial fraction decomposition is a useful technique that simplifies complex rational expressions. When you're faced with a fraction where both the numerator and the denominator are polynomials, partial fraction decomposition can split it into simpler "partial fractions." This makes it easier to work with, especially for integration or summation.
For instance, consider the general term in the given series, \( \frac{1}{k(k+1)} \). This fraction can seem daunting at first, but by applying partial fraction decomposition, it becomes more manageable. We express it as:\[\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\]This expression is equivalent; we can check this by finding a common denominator and simplifying. Decomposition helps reveal hidden patterns, like telescoping, that make solving problems easier. Understanding this breakdown is key to solving various calculus exercises efficiently.

Mathematical Series

Mathematical series are essentially the sum of terms in a sequence. Each term in the sequence has its own position and rules that define its value. In many math problems, particularly those in calculus, series play a crucial role. The given problem concerns itself with a specific type of series, where each term is structured as \( \frac{1}{n(n+1)} \).
This example showcases how series can be manipulated and simplified. By applying the decomposed form \( \frac{1}{k} - \frac{1}{k+1} \) to each term, the once complex series becomes straightforward.

• The series starts with \( \frac{1}{1 \cdot 2} \) and ends with \( \frac{1}{n(n+1)} \).
• By rewriting each term, we transform the series, allowing it to simplify naturally via telescoping.

Recognizing series configurations and their behavior, like geometric or arithmetic sequences, sharpens one’s ability to tackle calculus exercises efficiently.

Calculus Exercises

In calculus exercises, breaking down expressions into their simplest form is a common strategy. It's like peeling layers off an onion to get to the core problem-solving elements. This approach aligns with the exercise of finding a formula for the sum \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} \).
The key steps involve:

• Identifying the general term \( \frac{1}{k(k+1)} \),
• Applying partial fraction decomposition to simplify each term,
• Recognizing the telescoping nature of the series to drastically reduce the number of terms to consider, and finally,
• Calculating the remaining terms to find that the sum simplifies neatly to \( 1 - \frac{1}{n+1} \).

This exercise not only illustrates clever algebraic manipulation but also highlights the importance of recognizing patterns within calculus exercises that simplify complex expressions into elegant solutions.