Problem 9
Question
Determine whether the geometric series converges or diverges. If it converges, find its sum. \(\frac{5}{3}-\frac{5}{9}+\frac{5}{27}-\frac{5}{81}+\cdots\)
Step-by-Step Solution
Verified Answer
The given geometric series converges, since the common ratio \(r = -\frac{1}{3}\) lies between -1 and 1. The sum of the series can be found using the formula \(S = \frac{a_1}{1 - r}\), where \(a_1 = \frac{5}{3}\). Thus, the sum of the series is \(S = \frac{5}{4}\).
1Step 1: Identify the geometric series
The given series is:
\(\frac{5}{3}-\frac{5}{9}+\frac{5}{27}-\frac{5}{81}+\cdots\)
2Step 2: Identify the common ratio
To find the common ratio, we can divide any term by its previous term. Let's divide the second term by the first term:
\(r = \frac{-\frac{5}{9}}{\frac{5}{3}} = -\frac{1}{3}\)
So, the common ratio (\(r\)) of this series is -1/3.
3Step 3: Determine if the series converges or diverges
A geometric series converges if the common ratio is between -1 and 1. In this case, the common ratio is -1/3, which is between -1 and 1. Therefore, the series converges.
4Step 4: Find the sum of the series
If the geometric series converges, we can find its sum using the formula:
\(S = \frac{a_1}{1 - r}\)
where \(S\) is the sum of the series, \(a_1\) is the first term, and \(r\) is the common ratio. In our case, \(a_1 = \frac{5}{3}\) and \(r = -\frac{1}{3}\).
Plugging these values into the formula, we get:
\(S = \frac{\frac{5}{3}}{1 - (-\frac{1}{3})} = \frac{\frac{5}{3}}{\frac{4}{3}}\)
Now, we can divide \(\frac{5}{3}\) by \(\frac{4}{3}\) to get the sum:
\(S = \frac{5}{4}\)
So, the sum of the given geometric series is \(\frac{5}{4}\).
Key Concepts
Common RatioConvergence of SeriesSum of Series
Common Ratio
In every geometric series, a crucial component is the common ratio. It is the factor by which you multiply each term to move to the next term in the series. To identify the common ratio, you can divide any term of the series by the previous term. For example, in the series \(\frac{5}{3} - \frac{5}{9} + \frac{5}{27} - \frac{5}{81} + \cdots\), you can take the second term \(\frac{-5}{9}\) and divide it by the first term \(\frac{5}{3}\).
This gives you:
\[ r = \frac{-\frac{5}{9}}{\frac{5}{3}} = -\frac{1}{3} \]
So, in this series, the common ratio \( r \) is \(-\frac{1}{3}\). Understanding the common ratio helps in determining the convergence of the series and calculating its sum. It's essential to understand if the common ratio is positive or negative, as this affects the series' shape and direction.
This gives you:
\[ r = \frac{-\frac{5}{9}}{\frac{5}{3}} = -\frac{1}{3} \]
So, in this series, the common ratio \( r \) is \(-\frac{1}{3}\). Understanding the common ratio helps in determining the convergence of the series and calculating its sum. It's essential to understand if the common ratio is positive or negative, as this affects the series' shape and direction.
Convergence of Series
The convergence of a geometric series refers to whether the series has a finite sum as it progresses to infinity. For a geometric series to converge, the absolute value of the common ratio must be less than one. In other words, \(-1 < r < 1\).
For the given series, our common ratio \( r \) is \(-\frac{1}{3}\). Since \(-\frac{1}{3}\) lies within this range, we can conclude that the series converges.
If the common ratio were outside of this range (for example, if \( |r| \geq 1 \)), the series would diverge, meaning it wouldn't have a finite sum but would rather expand indefinitely or oscillate.
For the given series, our common ratio \( r \) is \(-\frac{1}{3}\). Since \(-\frac{1}{3}\) lies within this range, we can conclude that the series converges.
If the common ratio were outside of this range (for example, if \( |r| \geq 1 \)), the series would diverge, meaning it wouldn't have a finite sum but would rather expand indefinitely or oscillate.
- Converges: Series will reach a finite sum.
- Diverges: Series does not have a finite sum.
Sum of Series
When a geometric series converges, you can find its sum using a simple formula:
\[ S = \frac{a_1}{1 - r} \] where \( S \) is the sum of the series, \( a_1 \) is the first term, and \( r \) is the common ratio.
For our series, \( a_1 = \frac{5}{3} \) and \( r = -\frac{1}{3} \). Plugging these into the formula gives:
\[ S = \frac{\frac{5}{3}}{1 - (-\frac{1}{3})} = \frac{\frac{5}{3}}{\frac{4}{3}} \]
When you perform the division \( \frac{5}{3} \div \frac{4}{3} \), you obtain \( \frac{5}{4} \). This result is the sum of the series, showing that even though the series has infinitely many terms, they sum up to a finite number when the series converges.
By using this formula, you can easily evaluate the sum of any convergent geometric series by simply identifying the first term and the common ratio.
\[ S = \frac{a_1}{1 - r} \] where \( S \) is the sum of the series, \( a_1 \) is the first term, and \( r \) is the common ratio.
For our series, \( a_1 = \frac{5}{3} \) and \( r = -\frac{1}{3} \). Plugging these into the formula gives:
\[ S = \frac{\frac{5}{3}}{1 - (-\frac{1}{3})} = \frac{\frac{5}{3}}{\frac{4}{3}} \]
When you perform the division \( \frac{5}{3} \div \frac{4}{3} \), you obtain \( \frac{5}{4} \). This result is the sum of the series, showing that even though the series has infinitely many terms, they sum up to a finite number when the series converges.
By using this formula, you can easily evaluate the sum of any convergent geometric series by simply identifying the first term and the common ratio.
Other exercises in this chapter
Problem 9
Use the Comparison Test to determine whether the series is convergent or divergent. \(\sum_{n=2}^{\infty} \frac{\ln n}{n}\)
View solution Problem 9
Find the radius of convergence and the interval of convergence of the power series. $$ \sum_{n=2}^{\infty} \frac{x^{n}}{\ln n} $$
View solution Problem 9
Determine whether the series converges or diverges. $$ \sum_{n=2}^{\infty} \frac{(-1)^{n} n}{\ln n} $$
View solution Problem 9
Determine whether the \(p\) -series is convergent or divergent. $$ \sum_{n=1}^{\infty} \frac{1}{n^{3}} $$
View solution