Problem 47
Question
Find the sum of the infinite geometric series. $$ 1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots $$
Step-by-Step Solution
Verified Answer
The sum of the infinite series is \( \frac{3}{2} \).
1Step 1: Identify the First Term
In a geometric series, the first term is denoted by \(a\). For the given series \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots\), the first term \(a\) is 1.
2Step 2: Determine the Common Ratio
The common ratio \(r\) in a geometric series is found by dividing any term by the preceding term. For the second term, \(\frac{1}{3}\), divided by the first term, \(1\), the common ratio \(r = \frac{1}{3} \).
3Step 3: Check for Convergence
The sum of an infinite geometric series converges if the absolute value of the common ratio \(|r| < 1\). Here, \(|\frac{1}{3}| = \frac{1}{3} < 1\), so the series does converge.
4Step 4: Apply the Infinite Series Sum Formula
For an infinite geometric series, the sum \(S\) is given by the formula \( S = \frac{a}{1 - r} \). Substituting the known values, we have \( S = \frac{1}{1 - \frac{1}{3}} \).
5Step 5: Simplify to Find the Sum
Calculate the denominator: \(1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}\). Using the fraction division, \( \frac{1}{\frac{2}{3}} = \frac{3}{2} \). Therefore, the sum of the infinite series is \( \frac{3}{2} \).
Key Concepts
ConvergenceCommon RatioInfinite Series Sum Formula
Convergence
In the world of infinite geometric series, *convergence* is a vital concept.
When we talk about a series "converging," we're essentially looking at whether the series approaches a certain value as the number of terms increases indefinitely.An infinite series converges if the absolute value of its common ratio, denoted as \(|r|\), is less than 1. This condition ensures that each term in the series becomes progressively smaller, bringing the sum of the series closer to a finite limit.
In the example series given, \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\), the common ratio is \(\frac{1}{3}\). Since \(|\frac{1}{3}| = \frac{1}{3} < 1\), the series converges.Convergence plays a foundational role in determining if we can find the sum of an infinite series.
If the series did not converge, the sum would not be finite or calculable using normal methods.
When we talk about a series "converging," we're essentially looking at whether the series approaches a certain value as the number of terms increases indefinitely.An infinite series converges if the absolute value of its common ratio, denoted as \(|r|\), is less than 1. This condition ensures that each term in the series becomes progressively smaller, bringing the sum of the series closer to a finite limit.
In the example series given, \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\), the common ratio is \(\frac{1}{3}\). Since \(|\frac{1}{3}| = \frac{1}{3} < 1\), the series converges.Convergence plays a foundational role in determining if we can find the sum of an infinite series.
If the series did not converge, the sum would not be finite or calculable using normal methods.
Common Ratio
The *common ratio* is a key component of geometric series. It's the constant factor between consecutive terms.
For any geometric series, you can find the common ratio by dividing any term by the previous term.Consider the series provided: \(1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots\). To find the common ratio \(r\), we can divide the second term by the first term:
\(r = \frac{1}{3} \div 1 = \frac{1}{3}\).The common ratio gives us a clear picture of how quickly or slowly the terms of the series decrease or increase.
If \(r\) is positive and less than 1, like in our series, each term is smaller than the one before, leading to convergence.
For any geometric series, you can find the common ratio by dividing any term by the previous term.Consider the series provided: \(1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots\). To find the common ratio \(r\), we can divide the second term by the first term:
\(r = \frac{1}{3} \div 1 = \frac{1}{3}\).The common ratio gives us a clear picture of how quickly or slowly the terms of the series decrease or increase.
If \(r\) is positive and less than 1, like in our series, each term is smaller than the one before, leading to convergence.
Infinite Series Sum Formula
To find the sum of an infinite geometric series quickly, we use the *infinite series sum formula*:
\[ S = \frac{a}{1 - r} \]
where \(S\) is the sum of the series, \(a\) is the first term, and \(r\) is the common ratio. This formula only works when the series converges, which happens when \(|r| < 1\).For the series \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\), we identified that \(a = 1\) and \(r = \frac{1}{3}\).
Plugging these values into the formula results in:
It's a powerful tool in simplifying complex series into a singular comprehensible value.
\[ S = \frac{a}{1 - r} \]
where \(S\) is the sum of the series, \(a\) is the first term, and \(r\) is the common ratio. This formula only works when the series converges, which happens when \(|r| < 1\).For the series \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\), we identified that \(a = 1\) and \(r = \frac{1}{3}\).
Plugging these values into the formula results in:
- \( S = \frac{1}{1 - \frac{1}{3}} \)
- \( S = \frac{1}{\frac{2}{3}} \)
- This simplifies down to \( S = \frac{3}{2} \)
It's a powerful tool in simplifying complex series into a singular comprehensible value.
Other exercises in this chapter
Problem 46
Find the sum. $$\sum_{i=1}^{3} i 2^{i}$$
View solution Problem 47
Show that \(\left(\begin{array}{l}{n} \\\ {1}\end{array}\right)=\left(\begin{array}{c}{n} \\ {n-1}\end{array}\right)=n\)
View solution Problem 47
Use a graphing calculator to evaluate the sum. $$\sum_{k=1}^{10} k^{2}$$
View solution Problem 48
Show that \(\left(\begin{array}{l}{n} \\\ {r}\end{array}\right)=\left(\begin{array}{c}{n} \\ {n-r}\end{array}\right) \quad\) for \(0 \leq r \leq n\)
View solution