Chapter 4: Problem 1
Find the sum of \(\sum_{n=3}^{\infty} \frac{2^{n}}{e^{n+1}}\).
Short Answer
Expert verified
The sum is \( \frac{8}{e^3(e-2)} \).
Step by step solution
01
Identify the Series Type
The series in question is \( \sum_{n=3}^{\infty} \frac{2^n}{e^{n+1}} \). This is an infinite geometric series because each term is obtained by multiplying the previous term by a constant ratio.
02
Find the First Term
To find the first term of the series, plug \( n = 3 \) into the series formula: \( \frac{2^3}{e^{3+1}} = \frac{8}{e^4} \). Thus, the first term, \( a \), is \( \frac{8}{e^4} \).
03
Determine the Common Ratio
The common ratio \( r \) can be found by dividing the second term by the first term. The second term is obtained by plugging \( n = 4 \) into the series formula, giving \( \frac{2^4}{e^{4+1}} = \frac{16}{e^5} \). Hence, the common ratio \( r = \frac{16/e^5}{8/e^4} = \frac{2}{e} \).
04
Check for Convergence
For the series to converge, the absolute value of the common ratio \( r \) must be less than 1. Since \( |r| = \frac{2}{e} \) and \( e \approx 2.718 \), it follows that \( \frac{2}{e} < 1 \). The series converges.
05
Calculate the Sum
The sum of an infinite geometric series is given by the formula \( S = \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio. Substituting the values from previous steps gives \( S = \frac{8/e^4}{1 - \frac{2}{e}} \). Simplify the expression to get \( S = \frac{8/e^4}{\frac{e-2}{e}} \). This further simplifies to \( S = \frac{8}{e^3(e-2)} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Infinite Series
An infinite series is a sum of terms that continues indefinitely. In mathematics, it is often expressed as \( \sum_{n=1}^{\infty} a_n \), where each \( a_n \) represents a term in the series. The beauty of infinite series lies in their ability to converge to a specific value, despite having an infinite number of terms.
Infinite series are prevalent in various fields, from calculus to physics, and they allow the representation of functions or numbers in a compact form. Here's a quick breakdown to help understand infinite series further:
Infinite series are prevalent in various fields, from calculus to physics, and they allow the representation of functions or numbers in a compact form. Here's a quick breakdown to help understand infinite series further:
- An infinite arithmetic series involves a constant addition to each successive term.
- An infinite geometric series, like the exercise presented, involves multiplying each term by a constant ratio to get to the next term.
Convergence of Series
The concept of convergence is central to the study of infinite series. For a series to be considered convergent, its sum must approach a finite limit as more terms are added. When dealing with infinite geometric series, the series will converge if the absolute value of the common ratio \( r \) is less than 1.
In mathematical terms, convergence is described by:
In mathematical terms, convergence is described by:
- \( |r| < 1 \) for convergence of a geometric series.
- \( |r| \geq 1 \) means the series will diverge and not have a finite sum.
Sum of Series
Once convergence is established for an infinite series, calculating its sum becomes possible. For infinite geometric series, there's a handy formula available:\[S = \frac{a}{1-r}\]where \( S \) is the sum, \( a \) is the first term, and \( r \) is the common ratio. This formula simplifies the process of finding the sum when the series converges.
In the exercise, using the identified first term \( \frac{8}{e^4} \) and common ratio \( \frac{2}{e} \), we incorporated these into the formula to find the sum:
In the exercise, using the identified first term \( \frac{8}{e^4} \) and common ratio \( \frac{2}{e} \), we incorporated these into the formula to find the sum:
- \( S = \frac{8/e^4}{1 - \frac{2}{e}} \)
- This further simplifies to \( S = \frac{8}{e^3(e-2)} \)