Problem 10
Question
Verify that the infinite series diverges. $$ \sum_{n=0}^{\infty} 2(-1.03)^{n} $$
Step-by-Step Solution
Verified Answer
The infinite geometric series \(\sum_{n=0}^{\infty} 2(-1.03)^{n}\) diverges since the absolute value of the ratio (1.03) is greater than 1.
1Step 1: Identify the Series Type
The series provided can be written in the form of a geometric series, which is \(\sum a*r^{n}\). Here, \(a = 2\) and \(r = -1.03\).
2Step 2: Evaluate the Ratio
The constant ratio for this geometric series is \(r = -1.03\).
3Step 3: Apply the Convergence Condition
A geometric series \(\sum ar^{n}\) is convergent if \(-1 < r < 1\) and divergent otherwise. Since the \(r\) value is greater than 1 or less than -1, the series is divergent.
Key Concepts
Geometric SeriesConvergence ConditionDivergent Series
Geometric Series
A geometric series is a type of series where each term is derived from the previous one by multiplying it by a constant factor, called the "common ratio." This kind of series follows the form \(\sum a \cdot r^{n}\), where \(a\) represents the first term of the series, \(r\) is the common ratio, and \(n\) starts from 0 and goes to infinity.
A classical example of a geometric series is \(2 + 2r + 2r^2 + 2r^3 + \ldots\).
Things to remember about geometric series:
A classical example of a geometric series is \(2 + 2r + 2r^2 + 2r^3 + \ldots\).
Things to remember about geometric series:
- The first term, denoted as \(a\), remains constant throughout the series.
- The common ratio \(r\) determines the behavior and pattern of the series.
- Understanding these two components helps greatly in determining whether the series converges or diverges.
Convergence Condition
The convergence of a geometric series is dependent on its common ratio. The rule is quite simple: a geometric series converges if the absolute value of the common ratio \(r\) is less than 1, that is, \(-1 < r < 1\).
If the series converges, it approaches a finite limit as more terms are added. The sum of such an infinite geometric series can be calculated as \(S = \frac{a}{1-r}\), where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio.
However, if \(r\) falls outside of this range, the series does not settle around a specific number and is known as divergent.
If the series converges, it approaches a finite limit as more terms are added. The sum of such an infinite geometric series can be calculated as \(S = \frac{a}{1-r}\), where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio.
- A value of \(|r| < 1\) leads to convergence and an overall finite sum.
- It’s a quick and efficient way to determine the behavior of the series.
However, if \(r\) falls outside of this range, the series does not settle around a specific number and is known as divergent.
Divergent Series
When a series is classified as divergent, it means that as you add an infinite number of terms, there is no fixed sum. In other words, the series doesn't stabilize to a finite value as more and more terms are added. A geometric series with a common ratio \(r\) that is equal to or greater than 1, or less than -1, will always be divergent.
For this specific exercise, the series \(\sum_{n=0}^{\infty} 2(-1.03)^{n}\) is divergent since the common ratio \(r = -1.03\) falls outside the convergence range \(-1 < r < 1\).
Key points about divergent series include:
Hence, understanding the divergence of a series is crucial in analysis and helps predict the behavior of series when summing infinitely many terms.
For this specific exercise, the series \(\sum_{n=0}^{\infty} 2(-1.03)^{n}\) is divergent since the common ratio \(r = -1.03\) falls outside the convergence range \(-1 < r < 1\).
Key points about divergent series include:
- The series does not sum to a single, finite limit.
- The value of \(r\) determines the divergence - in this case, being greater than 1 in absolute value leads to divergence.
Hence, understanding the divergence of a series is crucial in analysis and helps predict the behavior of series when summing infinitely many terms.
Other exercises in this chapter
Problem 9
Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \sin \frac{(2 n-1) \pi}{2} $$
View solution Problem 9
Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ f(x)=\frac{3}{x+2}, \quad c=0 $$
View solution Problem 10
Write the first five terms of the recursively defined sequence. \(a_{1}=6, a_{k+1}=\frac{1}{3} a_{k}^{2}\)
View solution Problem 10
Use the Integral Test to determine the convergence or divergence of the series. $$ \sum_{n=2}^{\infty} \frac{1}{n \sqrt{\ln n}} $$
View solution