Chapter 8: Problem 28
Determine whether the series is convergent or divergent by expressing \(s_{n}\) as a telescoping sum (as in Example 6\() .\) If it is convergent, find its sum. $$\sum_{n=1}^{\infty}\left(e^{1 / n}-e^{1 /(n+1)}\right)$$
Short Answer
Expert verified
The series is convergent, with sum \( e - 1 \).
Step by step solution
01
Identify the Series Type
The series given is \( \sum_{n=1}^{\infty}\left(e^{1/n} - e^{1/(n+1)}\right) \). Notice that this is structured as a difference of consecutive terms, which suggests it could be a telescoping series.
02
Write the General Term Difference
The general term of the series is the difference \( e^{1/n} - e^{1/(n+1)} \). Rewriting this shows that as \( n \) increases, terms can cancel out with those in preceding or succeeding summands.
03
Set Up the Partial Sum
To see the telescoping nature, consider the partial sum \( s_k = \sum_{n=1}^{k} \left( e^{1/n} - e^{1/(n+1)} \right) \). This helps visualize cancellation.
04
Simplify the Telescoping Sum
The terms in the sequence cancel such that \( s_k = (e^{1/1} - e^{1/(k+1)}) \). This is because all intermediate exponential terms cancel out in the summation.
05
Take the Limit of the Partial Sums
The series sum \( S \) can be found by taking the limit as \( k \to \infty \), of the remaining terms in \( s_k \) which is \( S = \lim_{k \to \infty} (e^{1} - e^{1/(k+1)}) \).
06
Evaluate the Limit
Evaluate the limit, noting that as \( k \to \infty \), \( e^{1/(k+1)} \to 1 \). Therefore, the sum of the series is \( e - 1 \).
07
Final Step: Determine Convergence and Sum
Since the limit \( S = e - 1 \) is finite, the series is convergent, and the sum of the series is \( e - 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series Convergence
Series convergence is a core concept in mathematical analysis. It tells us if the sum of the terms in an infinite series approaches a specific, finite value. To determine if a series converges, we often look at its structure.
In the given problem, the series is expressed as \( \sum_{n=1}^{\infty} \left(e^{1/n} - e^{1/(n+1)}\right) \). Since it is set up as a difference of terms, it is a prime candidate for a telescoping series—where many terms cancel out with each other.
To check for convergence, we often use partial sums, denoted here by \( s_k \). This means we look at the sum of the first \( k \) terms to see what happens as \( k \) approaches infinity. If the partial sum approaches a finite number, then the series converges. For example, in the given exercise, the sum tends to \( e - 1 \) as \( k \to \infty \), indicating convergence.
In the given problem, the series is expressed as \( \sum_{n=1}^{\infty} \left(e^{1/n} - e^{1/(n+1)}\right) \). Since it is set up as a difference of terms, it is a prime candidate for a telescoping series—where many terms cancel out with each other.
To check for convergence, we often use partial sums, denoted here by \( s_k \). This means we look at the sum of the first \( k \) terms to see what happens as \( k \) approaches infinity. If the partial sum approaches a finite number, then the series converges. For example, in the given exercise, the sum tends to \( e - 1 \) as \( k \to \infty \), indicating convergence.
Partial Sums
Partial sums are a helpful tool used to understand the behavior of an entire series by looking at the sum of its initial terms. When analyzing series, we often define \( s_k = \sum_{n=1}^{k} a_n \), where \( a_n \) is the general term of our series.
Visualizing partial sums can be particularly useful when dealing with telescoping series - a series where terms cancel out each other over sequential sums. In the series \( \sum_{n=1}^{\infty} \left(e^{1/n} - e^{1/(n+1)}\right) \), the partial sum \( s_k \) simplifies as more terms are added. Here, the sum turns into \( e^{1} - e^{1/(k+1)} \), where most terms wipe out due to telescoping, reducing complexity and paving the way to easy evaluation for convergence.
Ultimately, by taking the limit of these partial sums as \( k \to \infty \), we can conclude whether the sum converges and, thus, find the overall sum of the series.
Visualizing partial sums can be particularly useful when dealing with telescoping series - a series where terms cancel out each other over sequential sums. In the series \( \sum_{n=1}^{\infty} \left(e^{1/n} - e^{1/(n+1)}\right) \), the partial sum \( s_k \) simplifies as more terms are added. Here, the sum turns into \( e^{1} - e^{1/(k+1)} \), where most terms wipe out due to telescoping, reducing complexity and paving the way to easy evaluation for convergence.
Ultimately, by taking the limit of these partial sums as \( k \to \infty \), we can conclude whether the sum converges and, thus, find the overall sum of the series.
Exponential Function
The exponential function, denoted by \( e^x \), is a crucial building block in mathematics, particularly known for its unique properties in calculus and exponential growth scenarios. Here, \( e \approx 2.71828 \), is the base of natural logarithms.
In the given series problem, the terms \( e^{1/n} \) and \( e^{1/(n+1)} \) are used. These exponential expressions are used in the structure of the telescoping series, highlighting the function's role in easing complex problem-solving.
The nature of the exponential function allows for simplifications due to its continuous and differentiable properties, which is handy when terms are structured in a way that can telescope. This often results in mathematical elegance, as seen where intermediate terms vanish, dramatically simplifying the resultant sums.
In the given series problem, the terms \( e^{1/n} \) and \( e^{1/(n+1)} \) are used. These exponential expressions are used in the structure of the telescoping series, highlighting the function's role in easing complex problem-solving.
The nature of the exponential function allows for simplifications due to its continuous and differentiable properties, which is handy when terms are structured in a way that can telescope. This often results in mathematical elegance, as seen where intermediate terms vanish, dramatically simplifying the resultant sums.