Problem 15
Question
In Exercises \(15-18,\) determine if the geometric series converges or diverges. If a series converges, find its sum. $$ 1+\left(\frac{2}{5}\right)+\left(\frac{2}{5}\right)^{2}+\left(\frac{2}{5}\right)^{3}+\left(\frac{2}{5}\right)^{4}+\cdots $$
Step-by-Step Solution
Verified Answer
The series converges with a sum of \(\frac{5}{3}\).
1Step 1: Identify the Geometric Series Parameters
First, recognize that the series is in the form of a geometric series. The general form of a geometric series is \(a + ar + ar^2 + ar^3 + \cdots \). In this particular series, \(a = 1\) and \(r = \frac{2}{5}\).
2Step 2: Determine Convergence or Divergence
A geometric series converges if the absolute value of the common ratio \(|r|\) is less than 1. Here, \(r = \frac{2}{5}\) and \(|r| = \frac{2}{5} < 1\). Therefore, this series converges.
3Step 3: Calculate the Sum of the Convergent Series
For a convergent geometric series, the sum \(S\) can be calculated using the formula: \[ S = \frac{a}{1-r} \] Substitute \(a = 1\) and \(r = \frac{2}{5}\) into the formula:\[S = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}\]
4Step 4: Provide Final Answer
Having verified that the series converges and calculated its sum, the geometric series converges with a sum of \(\frac{5}{3}\).
Key Concepts
Series ConvergenceGeometric SequenceSum of a Series
Series Convergence
Series convergence is an important concept in mathematics when examining infinite series. It helps us determine if adding an infinite number of terms will yield a finite sum. In general, a series converges when the sum of its terms approaches a specific value.
For the given series: \(1 + \left(\frac{2}{5}\right) + \left(\frac{2}{5}\right)^2 + \left(\frac{2}{5}\right)^3 + \cdots\), the common ratio is \(r = \frac{2}{5}\). Since \(|\frac{2}{5}| < 1\), we know this series converges.
- A series is divergent if the sum does not approach any limit and instead heads toward infinity.
- To check for convergence, we often look at the behavior of the series’ terms as the series progresses.
For the given series: \(1 + \left(\frac{2}{5}\right) + \left(\frac{2}{5}\right)^2 + \left(\frac{2}{5}\right)^3 + \cdots\), the common ratio is \(r = \frac{2}{5}\). Since \(|\frac{2}{5}| < 1\), we know this series converges.
Geometric Sequence
Understanding a geometric sequence is an essential step when working with a geometric series. A geometric sequence is a sequence of numbers where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio.
The series in this exercise follows the form of a geometric sequence:
The series in this exercise follows the form of a geometric sequence:
- The first term (\(a\)) is \(1\), and represents the starting point of the series.
- The second term is \(ar\), where \(r = \frac{2}{5}\).
- This pattern continues with terms \(ar^2, ar^3, ar^4,\ldots\).
Sum of a Series
The sum of a series is perhaps the most intriguing part for those studying sequences and series. For a converging geometric series, there is a specific formula used to find its sum:
- First term \(a = 1\)
- Common ratio \(r = \frac{2}{5}\)
Substituting these values results in:\[ S = \frac{1}{1-\frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]This calculation shows that the sum of the infinite series is \(\frac{5}{3}\). Recognizing the summed value allows us to understand complex problems and apply these concepts to real-world mathematics.
- The formula \( S = \frac{a}{1-r} \) allows for a quick calculation of the infinite series' sum once convergence is confirmed.
- Substitute the values of the first term \(a\) and the common ratio \(r\) into this formula to get the sum.
- First term \(a = 1\)
- Common ratio \(r = \frac{2}{5}\)
Substituting these values results in:\[ S = \frac{1}{1-\frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]This calculation shows that the sum of the infinite series is \(\frac{5}{3}\). Recognizing the summed value allows us to understand complex problems and apply these concepts to real-world mathematics.
Other exercises in this chapter
Problem 15
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