Problem 41
Question
Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{2^{n}-n}{n 2^{n}} $$
Step-by-Step Solution
Verified Answer
The series diverges.
1Step 1: Simplify the Series Expression
Start by simplifying the given series's general term \( \frac{2^n - n}{n 2^n} \). You can rewrite it as \( \frac{2^n}{n 2^n} - \frac{n}{n 2^n} \), which simplifies to \( \frac{1}{n} - \frac{1}{2^n} \). Hence, the series can be rewritten as \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{2^n} \right) \).
2Step 2: Analyze the Series
Notice that the simplified series \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{2^n} \right) \) consists of two separate series: \( \sum_{n=1}^{\infty} \frac{1}{n} \) and \( \sum_{n=1}^{\infty} \frac{1}{2^n} \). The series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is a harmonic series, known to diverge. The series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) is a geometric series with a common ratio of \( \frac{1}{2} \), which converges.
3Step 3: Determine Convergence or Divergence
Since the series is the difference of a diverging series (harmonic series) and a converging geometric series, the overall behavior will be dominated by the diverging series. Therefore, \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{2^n} \right) \) diverges because subtracting a converging series from a diverging one does not stop the series from diverging.
Key Concepts
Geometric SeriesHarmonic SeriesInfinite SeriesSeries Simplification
Geometric Series
A geometric series is a series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In mathematical terms, a geometric series can be represented as \(\sum_{n=0}^{\infty} ar^n\), where \(a\) is the first term and \(r\) is the common ratio.
One of the fundamental properties of a geometric series is that it converges if the absolute value of the common ratio \(r\) is less than 1. In such cases, the sum of the series can be calculated using the formula:
\[ S = \frac{a}{1-r} \]
If \(|r|\geq 1\), the series diverges, meaning it does not approach a finite limit.
In the example provided in the original exercise, the series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) is an example of a geometric series with a common ratio \(r = \frac{1}{2}\). Since \(|r|<1\), this series converges.
One of the fundamental properties of a geometric series is that it converges if the absolute value of the common ratio \(r\) is less than 1. In such cases, the sum of the series can be calculated using the formula:
\[ S = \frac{a}{1-r} \]
If \(|r|\geq 1\), the series diverges, meaning it does not approach a finite limit.
In the example provided in the original exercise, the series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) is an example of a geometric series with a common ratio \(r = \frac{1}{2}\). Since \(|r|<1\), this series converges.
Harmonic Series
A harmonic series is a particular type of series that is important in mathematics. It takes the form \(\sum_{n=1}^{\infty} \frac{1}{n}\). This series is famous for its divergence, even though the terms get smaller as \( n \) increases.
The divergence of the harmonic series means that the sum grows without bound and cannot be assigned to a finite number. Although each individual term decreases as \(n\) increases, the pace of decrease isn't quick enough to sum to a finite value.
In the solution provided, the series \(\sum_{n=1}^{\infty} \frac{1}{n}\) represents the harmonic portion of the problem. This characteristic is pivotal in determining that the overall series diverges, despite the subtraction of the convergent geometric series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\).
The divergence of the harmonic series means that the sum grows without bound and cannot be assigned to a finite number. Although each individual term decreases as \(n\) increases, the pace of decrease isn't quick enough to sum to a finite value.
In the solution provided, the series \(\sum_{n=1}^{\infty} \frac{1}{n}\) represents the harmonic portion of the problem. This characteristic is pivotal in determining that the overall series diverges, despite the subtraction of the convergent geometric series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\).
Infinite Series
Infinite series arise when we consider sums of infinitely many terms, typically indexed by natural numbers. The sum is written as \(\sum_{n=1}^{\infty} a_n\), where \(a_n\) represents the general term. Determining whether such a series converges (approaches a specific value) or diverges (does not settle on a value) is a central problem in analysis.
Analyzing convergence involves strategies such as the ratio test, root test, and direct comparison test, among others. Moreover, infinite series are foundational to many aspects of mathematics, including calculus where they facilitate the understanding of functions and their behaviors.
In the exercise at hand, the infinite series \(\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{2^n} \right)\) showcases the delicate balance required to assess convergence through combination of series types. Here, while one component of the series is convergent and another divergent, it's crucial to understand the dominating behavior through component analysis.
Analyzing convergence involves strategies such as the ratio test, root test, and direct comparison test, among others. Moreover, infinite series are foundational to many aspects of mathematics, including calculus where they facilitate the understanding of functions and their behaviors.
In the exercise at hand, the infinite series \(\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{2^n} \right)\) showcases the delicate balance required to assess convergence through combination of series types. Here, while one component of the series is convergent and another divergent, it's crucial to understand the dominating behavior through component analysis.
Series Simplification
Series simplification is a powerful technique for understanding and analyzing series by breaking them into more manageable parts. By deconstructing a complex series into simpler components, each part can be easily recognized and analyzed.
In the given exercise, the expression \(\sum_{n=1}^{\infty} \frac{2^{n}-n}{n 2^{n}}\) is simplified by expressing it as \(\sum_{n=1}^{\infty} \left (\frac{1}{n} - \frac{1}{2^n} \right)\). This technique helps in applying known convergence tests to the simplified form.
Simplification often involves factoring, partial fraction decomposition, or separating terms. These methods clarify the nature of the series and facilitate the application of convergence and divergence tests. In this way, complicated series can be understood in terms of harmonic, geometric, or other fundamental types, making it easier to interpret their behavior.
In the given exercise, the expression \(\sum_{n=1}^{\infty} \frac{2^{n}-n}{n 2^{n}}\) is simplified by expressing it as \(\sum_{n=1}^{\infty} \left (\frac{1}{n} - \frac{1}{2^n} \right)\). This technique helps in applying known convergence tests to the simplified form.
Simplification often involves factoring, partial fraction decomposition, or separating terms. These methods clarify the nature of the series and facilitate the application of convergence and divergence tests. In this way, complicated series can be understood in terms of harmonic, geometric, or other fundamental types, making it easier to interpret their behavior.
Other exercises in this chapter
Problem 41
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