Problem 3
Question
Suppose \(\Sigma a_{n}\) and \(\Sigma b_{n}\) are series with positive terms and
\(\Sigma b_{n}\) is known to be convergent.
(a) If \(a_{n}>b_{n}\) for all \(n,\) what can you say about \(\Sigma a_{n} ?\) Why?
(b) If \(a_{n}
Step-by-Step Solution
Verified Answer
(a) \(\Sigma a_n\) diverges; (b) \(\Sigma a_n\) converges.
1Step 1: Understanding Convergence and Divergence
For series with positive terms, convergence means the series sums to a finite value. If a series is convergent, then each term must get arbitrarily small but not necessarily zero. If a series is divergent, the series does not sum to a finite value.
2Step 2: Analyzing Case (a): a_n > b_n for all n
Given that \(a_n > b_n\) for all \(n\), and \(\Sigma b_n\) is convergent, then \(\Sigma a_n\) must be divergent. This is because a convergent series \(\Sigma b_n\) with each \(b_n\) less than a corresponding \(a_n\) implies that \(\Sigma a_n\) is larger and thus cannot converge to a finite value.
3Step 3: Analyzing Case (b): a_n < b_n for all n
For \(a_n < b_n\) and knowing \(\Sigma b_n\) is convergent, the series \(\Sigma a_n\) converges. This follows from the comparison test which states that if a series dominates another convergent series, the series with smaller terms is also convergent.
Key Concepts
Comparison TestConvergent SeriesDivergent Series
Comparison Test
The Comparison Test is a fundamental tool used to determine whether a series is convergent or divergent. It involves comparing the terms of a given series with the terms of another series whose convergence is already known. This test can be quite intuitive:
Applying this test can simplify the process of analyzing series, especially when dealing with series involving complex terms.
- If you have two series, \( \Sigma a_n \) and \( \Sigma b_n \), both with positive terms, and \( 0 \leq a_n \leq b_n \) for all \( n \), then if \( \Sigma b_n \) is convergent, \( \Sigma a_n \) must also be convergent.
- Conversely, if \( a_n \geq b_n \) for all \( n \), and \( \Sigma b_n \) diverges, \( \Sigma a_n \) will also diverge.
Applying this test can simplify the process of analyzing series, especially when dealing with series involving complex terms.
Convergent Series
A convergent series is one where the sum of its infinite terms approaches a finite limit. This concept signifies that as you add more and more terms, the incremental additions become negligible, and you tend towards a specific value.
For a series \( \Sigma a_n \) to be convergent:
For a series \( \Sigma a_n \) to be convergent:
- The terms \( a_n \) must get smaller and eventually become close to zero.
- Mathematically, for any given small number \( \epsilon > 0 \), there exists a positive integer \( N \) such that for all \( n > N \), the partial sums \( S = a_1 + a_2 + ... + a_n \) is within \( \epsilon \) of the total sum \( S \).
Divergent Series
A divergent series is one where the sums of its terms don't approach a finite value. Instead, it continues to grow without bounds as more terms are added. This may occur for several reasons: the terms might not get small enough fast enough, or their cumulative sum could essentially be infinite.
When dealing with series \( \Sigma a_n \):
When dealing with series \( \Sigma a_n \):
- Typically, if the terms \( a_n \) do not approach zero, the series will be divergent.
- Even if \( a_n \) approaches zero, this alone is not enough to ensure convergence. Sometimes more is going on with how terms add up than just individual term size.
Other exercises in this chapter
Problem 3
\(3-8=\) Test the series for convergence or divergence. $$ \frac{4}{7}-\frac{4}{8}+\frac{4}{9}-\frac{4}{10}+\frac{4}{11}-\cdots $$
View solution Problem 3
Find the radius of convergence and interval of convergence of the series. $$\sum_{n=1}^{\infty}(-1)^{n} n x^{n}$$
View solution Problem 3
List the first six terms of the sequence defined by $$a_{n}=\frac{n}{2 n+1}$$ Does the sequence appear to have a limit? If so, find it.
View solution Problem 4
Find a power series representation for the function and determine the interval of convergence. $$ f(x)=\frac{5}{1-4 x^{2}} $$
View solution