Problem 1
Question
In Exercises \(1-6,\) find a formula for the \(n\) th partial sum of each series and use it to find the series' sum if the series converges. $$ 2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\cdots+\frac{2}{3^{n-1}}+\cdots $$
Step-by-Step Solution
Verified Answer
The series converges and its sum is 3.
1Step 1: Recognize the Series Type
The series presented is a geometric series. In a geometric series, each term is obtained by multiplying the previous term by a constant, known as the common ratio. In this series, the first term is 2 and the common ratio is \(\frac{1}{3}\) because \(\frac{\frac{2}{3}}{2} = \frac{1}{3}\).
2Step 2: Express the General Term
For a geometric series with the first term \(a\) and common ratio \(r\), the general term can be expressed as:\[ a_n = a \cdot r^{n-1} \]For this series: \(a = 2\) and \(r = \frac{1}{3}\), so the general term \(a_n\) is:\[ a_n = 2 \cdot \left(\frac{1}{3}\right)^{n-1} \]
3Step 3: Find the Formula for the Partial Sum
The formula for the \(n\)th partial sum \(S_n\) of a geometric series is:\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \]Substitute the values \(a = 2\) and \(r = \frac{1}{3}\) into the formula:\[ S_n = 2 \cdot \frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}} \]Simplify the denominator: \(1 - \frac{1}{3} = \frac{2}{3}\), which gives:\[ S_n = 2 \cdot \frac{1 - \left(\frac{1}{3}\right)^n}{\frac{2}{3}} \]Further simplify the expression:\[ S_n = 3 \cdot \left(1 - \left(\frac{1}{3}\right)^n\right) \]
4Step 4: Determine if the Series Converges
A geometric series converges if the absolute value of the common ratio \(|r| < 1\). Here, \(r = \frac{1}{3}\), which satisfies \(|r| < 1\), indicating that the series converges.
5Step 5: Find the Sum of the Converging Series
If the series converges, the sum \(S\) of the infinite geometric series is given by:\[ S = \frac{a}{1 - r} \]Substitute \(a = 2\) and \(r = \frac{1}{3}\) into the formula:\[ S = \frac{2}{1 - \frac{1}{3}} \]Simplify the expression for the sum:\[ S = \frac{2}{\frac{2}{3}} \]Further simplify to obtain the series sum:\[ S = 3 \]
Key Concepts
Understanding Partial SumsThe Concept of ConvergenceCalculating the Series SumIdentifying the Common Ratio in a Series
Understanding Partial Sums
In the context of a geometric series, a partial sum represents the total of the first 'n' terms of the series. This is a crucial concept because it allows you to approximate the sum of the series by considering a finite number of terms, especially when dealing with infinite series. By calculating partial sums, you can understand how the series behaves as more terms are added.
For any geometric series, the partial sum, denoted as \(S_n\), is calculated using a specific formula. This formula is \(S_n = a \cdot \frac{1 - r^n}{1 - r}\), where 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms.
For any geometric series, the partial sum, denoted as \(S_n\), is calculated using a specific formula. This formula is \(S_n = a \cdot \frac{1 - r^n}{1 - r}\), where 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms.
- Helps in determining behavior for finite terms
- Allows insight into the full infinite series
- A valuable tool for series analysis
The Concept of Convergence
Convergence is a key concept in understanding series like geometric ones. For a series to converge, the absolute value of its common ratio must be less than 1, i.e., \(|r| < 1\). This means that as you continue to add more terms to the series, the partial sums get closer and closer to a certain value, called the series sum.
In the series from our exercise, the common ratio \(r\) is \(\frac{1}{3}\), which is indeed less than 1, ensuring convergence. Convergence indicates stability in the series and confirms that adding infinitely many terms won't blow up the sum to infinity, but instead, it settles at a fixed number.
In the series from our exercise, the common ratio \(r\) is \(\frac{1}{3}\), which is indeed less than 1, ensuring convergence. Convergence indicates stability in the series and confirms that adding infinitely many terms won't blow up the sum to infinity, but instead, it settles at a fixed number.
- Determines where the series will 'end' as more terms are added
- Ensures the series stays bounded and manageable
Calculating the Series Sum
The series sum is the value that the series converges to, if it does converge. Once we've confirmed convergence, finding the infinite series sum becomes straightforward with the formula \(S = \frac{a}{1 - r}\), where 'S' is the series sum.
Incorporating the given series:
Understanding the sum gives insight into how the infinite addition of terms solidifies into a single value.
Incorporating the given series:
- First term \(a = 2\)
- Common ratio \(r = \frac{1}{3}\)
- Series sum is calculated as \(S = \frac{2}{1 - \frac{1}{3}} = 3\)
Understanding the sum gives insight into how the infinite addition of terms solidifies into a single value.
Identifying the Common Ratio in a Series
The common ratio is a fundamental element of any geometric series. It dictates how each term relates to the previous one, defining the series' behavior over time. In mathematical terms, the common ratio 'r' is obtained by dividing any term in the series by its preceding term.
For our exercise, the common ratio is \(r = \frac{1}{3}\).
For our exercise, the common ratio is \(r = \frac{1}{3}\).
- Ensures consistent multiplicative progression of terms
- Vital for determining convergence
- Directly influences the series sum
Other exercises in this chapter
Problem 1
Use the Comparison Test to determine if each series converges or diverges. \begin{equation}\sum_{n=1}^{\infty} \frac{1}{n^{2}+30}\end{equation}
View solution Problem 1
Use the Integral Test to determine if the series in Exercises \(1-10\) converge or diverge. Be sure to check that the conditions of the Integral Test are satisf
View solution Problem 1
Each of Exercises \(1-6\) gives a formula for the \(n\) th term \(a_{n}\) of a sequence \(\left\\{a_{n}\right\\} .\) Find the values of \(a_{1}, a_{2}, a_{3},\)
View solution Problem 1
Use substitution (as in Example 4) to find the Taylor series at \(x=0\) of the functions in Exercises \(1-10\) . $$e^{-5 x}$$
View solution