Problem 8
Question
Verify that the infinite series diverges. $$ \sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n} $$
Step-by-Step Solution
Verified Answer
The infinite series \(\sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}\) diverges.
1Step 1: Identify the series as a geometric series
A geometric series takes the form \(\sum_{n=0}^{\infty} ar^{n}\), where \(a\) is the first term and \(r\) is the common ratio. In this series, \(a = 1\) and \(r = \frac{4}{3}\).
2Step 2: Apply the convergence test for geometric series
The convergence test states that the geometric series \(\sum_{n=0}^{\infty} ar^{n}\) converges if the absolute value of \(r\) is less than 1 and diverges if it is greater than or equal to 1. Since \(|\frac{4}{3}| > 1\), the series diverges.
Key Concepts
Understanding Geometric SeriesApplying the Convergence TestExploring Infinite Series
Understanding Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In the context of our exercise, we have the series \[\sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}\].
Here, the series starts with 1 (our first term, which is when \(n=0\)), and each subsequent term is \(\frac{4}{3}\) times the one before it. This creates a pattern of numbers that get increasingly larger because the common ratio is greater than 1, which is precisely \(\frac{4}{3}\) in this case. Through their very nature, geometric series can display very different behaviors depending on the value of their common ratio, leading us to the important concept of convergence and divergence.
Here, the series starts with 1 (our first term, which is when \(n=0\)), and each subsequent term is \(\frac{4}{3}\) times the one before it. This creates a pattern of numbers that get increasingly larger because the common ratio is greater than 1, which is precisely \(\frac{4}{3}\) in this case. Through their very nature, geometric series can display very different behaviors depending on the value of their common ratio, leading us to the important concept of convergence and divergence.
Applying the Convergence Test
To determine if a series is mathematically 'well-behaved' and settles towards a finite number, we use a convergence test. One particularly useful test for geometric series states that the series \[\sum_{n=0}^{\infty} ar^{n}\] converges (meaning it approaches a finite value) if and only if the absolute value of the common ratio \( r \) is less than 1, symbolically \(|r| < 1\).
This test is pivotal because it gives us a clear numerical criterion to classify the series into those that add up to a finite limit and those that grow without bounds. In the exercise we’re analyzing, since \(|\frac{4}{3}| > 1\), we apply this test and immediately see that the series does not meet the condition for convergence; it exceeds the boundary set by the test, which tells us the series will continue to grow infinitely.
This test is pivotal because it gives us a clear numerical criterion to classify the series into those that add up to a finite limit and those that grow without bounds. In the exercise we’re analyzing, since \(|\frac{4}{3}| > 1\), we apply this test and immediately see that the series does not meet the condition for convergence; it exceeds the boundary set by the test, which tells us the series will continue to grow infinitely.
Exploring Infinite Series
An infinite series is essentially the sum of an infinite sequence of terms. This concept is fascinating as it challenges our intuitive understanding that adding an infinite number of terms could result in anything finite. To make sense of it, mathematicians have devised criteria, like the convergence test, to categorize these series.
We encounter such series frequently in mathematics and applied sciences because they model processes that continue indefinitely. The divergence of the series in our exercise demonstrates that not all infinite processes lead to a finite sum. In many scenarios, if the conditions for convergence aren't met - as we saw when the absolute value of \( r \) was not less than 1 - the sum of the series cannot be a definite number and instead is said to 'diverge', or in layman's terms, it explodes to infinity without a cap.
We encounter such series frequently in mathematics and applied sciences because they model processes that continue indefinitely. The divergence of the series in our exercise demonstrates that not all infinite processes lead to a finite sum. In many scenarios, if the conditions for convergence aren't met - as we saw when the absolute value of \( r \) was not less than 1 - the sum of the series cannot be a definite number and instead is said to 'diverge', or in layman's terms, it explodes to infinity without a cap.
Other exercises in this chapter
Problem 7
Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n+1)}{\ln (n+1)} $$
View solution Problem 7
Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ g(x)=\frac{1}{2 x-5}, \quad c=-3 $$
View solution Problem 8
Write the first five terms of the sequence. \(a_{n}=10+\frac{2}{n}+\frac{6}{n^{2}}\)
View solution Problem 8
Use the Integral Test to determine the convergence or divergence of the series. $$ \sum_{n=2}^{\infty} \frac{\ln n}{n^{3}} $$
View solution