Problem 38
Question
Find the sum of the convergent series. $$ \sum_{n=0}^{\infty} 2\left(-\frac{2}{3}\right)^{n} $$
Step-by-Step Solution
Verified Answer
The sum of the series \(\sum_{n=0}^{\infty} 2(-2/3)^{n}\) is 1.2
1Step 1: Identify the values of a and r
Firstly, analyze the series. Here, the first term (\(a\)) is 2 and the common ratio (\(r\)) is \(-2/3\)
2Step 2: Substitute the values into the sum formula
The next step is to substitute the values of \(a\) and \(r\) into the sum formula of an infinite geometric series. The formula is \(S = \frac{a}{1 - r}\). So the sum \(S\) of the series is \(S = \frac{2}{1 - (-2/3)}\)
3Step 3: Calculate the sum
Solve the equation \(S = \frac{2}{1 - (-2/3)}\). Multiply the numerator and the denominator by 3 to get rid of the fraction in the denominator: \(S = \frac{2*3}{3 - (-2)} = \frac{6}{3 + 2} = 6/5=1.2\)
Key Concepts
Convergent SeriesCommon RatioSum Formula for Geometric Series
Convergent Series
When we talk about a convergent series, we're referring to a series where the sum of its infinite terms approaches a specific value as more terms are added. Unlike divergent series that grow without bounds, convergent series stabilize at a certain number. A classic example of a convergent series is the infinite geometric series, which has the form \( \sum_{n=0}^{\infty} ar^n \). Here, \( a \) represents the first term of the series and \( r \) the common ratio. The convergence largely depends on the value of \( r \). The series only converges if the absolute value of the common ratio \( |r| \) is less than 1.
In the given exercise, the series \( \sum_{n=0}^{\infty} 2 \left(-\frac{2}{3}\right)^n \) is convergent because the common ratio \( -\frac{2}{3} \) satisfies this condition. As the terms grow in number, the series converges to a specific sum, which can be determined using special formulas.
In the given exercise, the series \( \sum_{n=0}^{\infty} 2 \left(-\frac{2}{3}\right)^n \) is convergent because the common ratio \( -\frac{2}{3} \) satisfies this condition. As the terms grow in number, the series converges to a specific sum, which can be determined using special formulas.
Common Ratio
The common ratio is a crucial aspect of geometric series. It is the number that each term in the series is multiplied by to get the next term. For an infinite geometric series to converge, this ratio is vital. Specifically, the series converges if the absolute value of the common ratio \( |r| \) is strictly less than 1.
In our exercise example, the common ratio \( r \) is \(-\frac{2}{3}\). This indicates that every term in the series is obtained by multiplying the previous term by \(-\frac{2}{3}\). Since \(|-\frac{2}{3}| = \frac{2}{3} < 1\), the series is convergent and has a finite sum. Understanding the value and role of the common ratio is essential in determining whether a geometric series will converge and calculating its total sum.
In our exercise example, the common ratio \( r \) is \(-\frac{2}{3}\). This indicates that every term in the series is obtained by multiplying the previous term by \(-\frac{2}{3}\). Since \(|-\frac{2}{3}| = \frac{2}{3} < 1\), the series is convergent and has a finite sum. Understanding the value and role of the common ratio is essential in determining whether a geometric series will converge and calculating its total sum.
Sum Formula for Geometric Series
Calculating the sum of an infinite geometric series is simplified using a special formula. When the series converges, which happens when the common ratio \( |r| < 1 \), we use the formula \( S = \frac{a}{1 - r} \). Here, \( a \) is the first term, and \( r \) is the common ratio.
In the provided exercise, the first term \( a \) is 2, and the common ratio \( r \) is \(-\frac{2}{3}\). To find the sum \( S \), substitute these values into the sum formula: \( S = \frac{2}{1 - (-\frac{2}{3})} \). Solving this, we find \( S = \frac{2}{1 + \frac{2}{3}} \). To simplify, multiply the numerator and the denominator by 3, obtaining \( S = \frac{6}{3 + 2} = \frac{6}{5} = 1.2 \). Hence, the sum of this convergent series is 1.2, making it a perfect example of how the formula works.
In the provided exercise, the first term \( a \) is 2, and the common ratio \( r \) is \(-\frac{2}{3}\). To find the sum \( S \), substitute these values into the sum formula: \( S = \frac{2}{1 - (-\frac{2}{3})} \). Solving this, we find \( S = \frac{2}{1 + \frac{2}{3}} \). To simplify, multiply the numerator and the denominator by 3, obtaining \( S = \frac{6}{3 + 2} = \frac{6}{5} = 1.2 \). Hence, the sum of this convergent series is 1.2, making it a perfect example of how the formula works.
Other exercises in this chapter
Problem 37
Determine whether the series converges conditionally or absolutely, or diverges. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} $$
View solution Problem 37
Explain how to use the geometric series $$g(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}, \quad|x|
View solution Problem 38
Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{1 \cdot 3 \cdot 5
View solution Problem 38
In Exercises \(35-38,\) write an equivalent series with the index of summation beginning at \(n=1\). $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n+1}}{2 n+1} $$
View solution