Problem 32
Question
For the series given, determine how large \(n\) must be so that using the nth partial sum to approximate the series gives an error of no more than 0.0002. $$ \sum_{k=1}^{\infty} \frac{1}{k(k+1)} $$
Step-by-Step Solution
Verified Answer
Use the 4999th partial sum for an error of no more than 0.0002.
1Step 1: Simplify the Series
The series \( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \) can be simplified by partial fraction decomposition. We rewrite the term as \( \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \). This forms a telescoping series.
2Step 2: Identify the Pattern in Partial Sums
The partial sum \( S_n \) for this telescoping series \( S_n = \left( 1 - \frac{1}{n+1} \right) \). As \( n \to \infty \), \( S_n \to 1 \).
3Step 3: Calculate Remaining Error after Partial Sum
The error after the nth partial sum is given by the magnitude of the next term, which is \( \frac{1}{n+1} \). To ensure the error is no more than 0.0002, we set up the inequality \( \frac{1}{n+1} \leq 0.0002 \).
4Step 4: Solve for n
Rearrange the inequality \( \frac{1}{n+1} \leq 0.0002 \) to find \( n+1 \geq 5000 \). Solving for \( n \), we get \( n \geq 4999 \). Thus, \( n = 4999 \) is the smallest integer satisfying the condition.
Key Concepts
Partial SumTelescoping SeriesError EstimationPartial Fractions
Partial Sum
In mathematics, a "partial sum" is simply the sum of a part of a series. A series itself is an infinite sequence of numbers that are to be added together. When we calculate the first few terms, we are dealing with a partial sum. Calculating partial sums is helpful when dealing with series because it allows us to approximate the sum of an infinite series with a finite number of terms.
Consider a series \( S = a_1 + a_2 + a_3 + \ldots \), where \( a_i \) represents each term. The nth partial sum, denoted as \( S_n \), is the sum of the first \( n \) terms: \( S_n = a_1 + a_2 + a_3 + \ldots + a_n \).
The beauty of partial sums lies in approximation. Since an infinite series may not converge to a sum, or may be difficult to evaluate directly, finding its partial sums helps us understand the behavior of the series and approximate its value. The key aspect is the difference between the nth partial sum and the actual series sum, measured as error.
Consider a series \( S = a_1 + a_2 + a_3 + \ldots \), where \( a_i \) represents each term. The nth partial sum, denoted as \( S_n \), is the sum of the first \( n \) terms: \( S_n = a_1 + a_2 + a_3 + \ldots + a_n \).
The beauty of partial sums lies in approximation. Since an infinite series may not converge to a sum, or may be difficult to evaluate directly, finding its partial sums helps us understand the behavior of the series and approximate its value. The key aspect is the difference between the nth partial sum and the actual series sum, measured as error.
Telescoping Series
A "telescoping series" is a kind of series where most terms cancel out each other, making it easier to compute the sum of the series. This is possible through the nature of its terms being designed to cancel previous or next terms.
In the given problem, the series \( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \) is identified as a telescoping series. By using partial fraction decomposition, each term \( \frac{1}{k(k+1)} \) can be rewritten as \( \frac{1}{k} - \frac{1}{k+1} \).
Because of this beneficial property, when adding up the terms of a telescoping series, most of the intermediate terms vanish, leaving only a few terms from the beginning and end. For instance, the partial sum \( S_n \) is simplified to \( 1 - \frac{1}{n+1} \), showing that as \( n \) gets larger, the sum converges to 1.
This cancellation effect makes telescoping series an invaluable tool for simplifying complex expressions and easily finding approximations.
In the given problem, the series \( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \) is identified as a telescoping series. By using partial fraction decomposition, each term \( \frac{1}{k(k+1)} \) can be rewritten as \( \frac{1}{k} - \frac{1}{k+1} \).
Because of this beneficial property, when adding up the terms of a telescoping series, most of the intermediate terms vanish, leaving only a few terms from the beginning and end. For instance, the partial sum \( S_n \) is simplified to \( 1 - \frac{1}{n+1} \), showing that as \( n \) gets larger, the sum converges to 1.
This cancellation effect makes telescoping series an invaluable tool for simplifying complex expressions and easily finding approximations.
Error Estimation
"Error estimation" is crucial when dealing with series, especially when using partial sums to approximate their actual values. The error, in this context, is the difference between the truncated series (given by the partial sum) and the true value of the entire series.
For the exercise in question, the series \( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \) is well-suited to this due to its telescoping nature. The error after the nth partial sum, often denoted as \( E_n \), corresponds to the term immediately following the last term included in the partial sum: \( E_n = \frac{1}{n+1} \).
To keep the error within a desired threshold, such as 0.0002 in this problem, we develop an inequality: \( \frac{1}{n+1} \leq 0.0002 \). Solving this inequality gives us the smallest \( n \) for which the approximation is sufficiently accurate. Appropriate error estimation is vital in numerical methods where precision matters.
For the exercise in question, the series \( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \) is well-suited to this due to its telescoping nature. The error after the nth partial sum, often denoted as \( E_n \), corresponds to the term immediately following the last term included in the partial sum: \( E_n = \frac{1}{n+1} \).
To keep the error within a desired threshold, such as 0.0002 in this problem, we develop an inequality: \( \frac{1}{n+1} \leq 0.0002 \). Solving this inequality gives us the smallest \( n \) for which the approximation is sufficiently accurate. Appropriate error estimation is vital in numerical methods where precision matters.
Partial Fractions
"Partial fractions" play a pivotal role in simplifying certain kinds of algebraic fractions, allowing them to be broken into simpler, more manageable parts. This technique is especially useful in calculus when integrating rational functions or simplifying the terms of a series to identify patterns.
The original exercise uses partial fraction decomposition for the series term \( \frac{1}{k(k+1)} \). Here, it is expressed as \( \frac{1}{k} - \frac{1}{k+1} \).
Breaking down complex rational expressions into partial fractions simplifies calculations and reveals simplifying patterns like those in telescoping series. By rewriting into partial fractions, it allows for straightforward summation and easier error estimation.
This decomposition offers a clearer form that is easier to integrate, differentiate, or sum, showcasing why partial fractions are an essential tool in mathematical analysis.
The original exercise uses partial fraction decomposition for the series term \( \frac{1}{k(k+1)} \). Here, it is expressed as \( \frac{1}{k} - \frac{1}{k+1} \).
Breaking down complex rational expressions into partial fractions simplifies calculations and reveals simplifying patterns like those in telescoping series. By rewriting into partial fractions, it allows for straightforward summation and easier error estimation.
This decomposition offers a clearer form that is easier to integrate, differentiate, or sum, showcasing why partial fractions are an essential tool in mathematical analysis.
Other exercises in this chapter
Problem 32
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