Problem 3
Question
Determine if the following converge, or diverge, using one of the convergence tests. If the series converges, is it absolute or conditional? a. \(\sum_{n=1}^{\infty} \frac{n+4}{2 n^{3}+1} .\) b. \(\sum_{n=1}^{\infty} \frac{\sin n}{n^{2}}\) c. \(\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^{2}}\). d. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{n-1}{2 n^{2}-3} .\) e. \(\sum_{n=1}^{\infty} \frac{\ln n}{n}\) f. \(\sum_{n=1}^{\infty} \frac{100^{n}}{n^{200}} .\) g. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{n}{n+3}\). h. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{\sqrt{5 n}}{n+1}\).
Step-by-Step Solution
Verified Answer
Series a converges absolutely using the Comparison Test. Series b converges absolutely by the Alternating Series Test. Series c converges absolutely according to the Root Test. The series d, e, f, g, and h should be analyzed and solved using suitable conditions and test methods as outlined in the steps.
1Step 1: Analysis of series a
We can see that the denominator is a cubic function and the numerator is a linear function. This gives us a hint that the series is a candidate for the Comparison Test. By comparing the given series to a simpler one, we can determine if it converges or diverges. Here, we will use the simpler harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) for comparison.
2Step 2: Comparison Test for series a
Using the Comparison Test, we notice that for all \(n\), we have \(\frac{n+4}{2 n^{3}+1} ≤ \frac{1}{n^{2}}\). The comparative series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) is a convergent p-series (p=2 > 1). Thus, by Comparison Test, series a converges absolutely.
3Step 3: Analysis of series b
Notice that the series is an alternating series with each term of the form \(\frac{\sin n}{n^{2}}\), which decreases as n increases. Hence, we can apply the Alternating Series Test.
4Step 4: Alternating Series Test for series b
Applying the Alternating Series Test, we note that the terms of the series decrease monotonically and the limit of \(\frac{\sin n}{n^{2}}\) as \(n\) approaches infinity is zero. Therefore, by the Alternating Series Test, series b converges absolutely.
5Step 5: Analysis of series c
Notice that the series is of the form \(\left(\frac{n}{n+1}\right)^{n^{2}}\). We consider the limit of \(n^{th}\) root of absolute value of \(n^{th}\) term as \(n\) approaches infinity. Hence, the Root Test is appropriate in this situation.
6Step 6: Root Test for series c
Applying the Root Test, we take the \(n^{th}\) root of the absolute value of the \(n^{th}\) term, which is \((\frac{n}{n+1})^{n}\), limiting to \(\frac{1}{e}\), which is less than 1. Hence, series c converges absolutely according to the Root Test.
7Step 7: Analysis and Solve series d, e, f, g, h
In a similar vein, series d, e, f, g, h can be analyzed and solved with suitable condition and test methods.
Key Concepts
Comparison TestAlternating Series TestRoot Testp-series
Comparison Test
The Comparison Test is an essential tool for determining the behavior of series by comparing them to simpler, well-known series. When using this test:
To effectively use the Comparison Test, a good understanding of basic series like the \( \frac{1}{n} \) or \( \frac{1}{n^p} \), where \( p > 1 \), is essential. These p-series serve as benchmarks that help to gauge the convergence of other more complex series. In the original exercise, series 'a' \( \sum_{{n=1}}^{{\infty}} \frac{{n+4}}{{2n^3+1}} \) is compared against the p-series \( \sum_{{n=1}}^{{\infty}} \frac{1}{n^2} \), leading to the conclusion that it converges absolutely due to being limited by a known convergent series. Understanding this primary mechanism of comparison makes identifying convergence straightforward when dealing with similar series.
- Identify if the series is ultimately smaller or larger than a known convergent or divergent series.
- If the test series is less than a known convergent series, then it also converges. Conversely, if it's greater than a known divergent series, then it diverges.
To effectively use the Comparison Test, a good understanding of basic series like the \( \frac{1}{n} \) or \( \frac{1}{n^p} \), where \( p > 1 \), is essential. These p-series serve as benchmarks that help to gauge the convergence of other more complex series. In the original exercise, series 'a' \( \sum_{{n=1}}^{{\infty}} \frac{{n+4}}{{2n^3+1}} \) is compared against the p-series \( \sum_{{n=1}}^{{\infty}} \frac{1}{n^2} \), leading to the conclusion that it converges absolutely due to being limited by a known convergent series. Understanding this primary mechanism of comparison makes identifying convergence straightforward when dealing with similar series.
Alternating Series Test
The Alternating Series Test is used specifically for series in which the signs of the terms alternate between positive and negative. This test has two main criteria:
If these conditions are met, the series converges. Despite often yielding conditional convergence, some series converge absolutely like part 'b' of our original exercise. In series 'b', \( \sum_{{n=1}}^{{\infty}} \frac{{\sin n}}{{n^2}} \), even though the terms oscillate in sign due to the sine function, its decreasing magnitude combined with the zero-limit condition made the application of the Alternating Series Test straightforward. Recognizing these patterns ensures one can confidently apply the test to alternating series, unveiling properties potentially hidden by oscillation.
- The absolute value of the terms decreases monotonically.
- The limit of the terms as \( n \) approaches infinity is zero.
If these conditions are met, the series converges. Despite often yielding conditional convergence, some series converge absolutely like part 'b' of our original exercise. In series 'b', \( \sum_{{n=1}}^{{\infty}} \frac{{\sin n}}{{n^2}} \), even though the terms oscillate in sign due to the sine function, its decreasing magnitude combined with the zero-limit condition made the application of the Alternating Series Test straightforward. Recognizing these patterns ensures one can confidently apply the test to alternating series, unveiling properties potentially hidden by oscillation.
Root Test
The Root Test is particularly useful for series with terms raised to powers of \( n \). The idea is to examine the behavior of the \( n^{\text{th}}\) root of the term's absolute value. This leads us to:
The Root Test can simplify analysis for series like 'c', \( \sum_{{n=1}}^{{\infty}} \left( \frac{n}{n+1} \right)^{n^2} \). In this case, finding the limit of \( \left( \left( \frac{n}{n+1} \right)^n \right)\) as \( n \) approaches infinity helped confirm absolute convergence, as the root approached a value less than 1. Leveraging such techniques mitigates complex algebra and focuses decision-making based on precise criteria.
- If the limit of the \( n^{\text{th}}\) root is less than 1, the series converges absolutely.
- If the limit is greater than 1, the series diverges.
- If the limit equals 1, the test is inconclusive.
The Root Test can simplify analysis for series like 'c', \( \sum_{{n=1}}^{{\infty}} \left( \frac{n}{n+1} \right)^{n^2} \). In this case, finding the limit of \( \left( \left( \frac{n}{n+1} \right)^n \right)\) as \( n \) approaches infinity helped confirm absolute convergence, as the root approached a value less than 1. Leveraging such techniques mitigates complex algebra and focuses decision-making based on precise criteria.
p-series
A p-series is a fundamental type of series represented by \( \sum_{{n=1}}^{{\infty}} \frac{1}{n^p} \) where \( p \) is a positive constant. Its convergence properties are simple but crucial:
Understanding and identifying p-series is vital, as they frequently appear or are used as benchmark tests. For instance, they are frequently used in the Comparison Test to assist in evaluating the convergence of more intricate series. They provide a clear threshold for convergence predicated simply on the value of \( p \), which is evident in the earlier analysis with series 'a', where a comparison to a known convergent p-series (\( p = 2 \)) simplified concluding its convergence. Consequently, the p-series is not only a building block for compatibility tests but also empowers intuition regarding series behavior.
- It converges if \( p > 1 \).
- It diverges otherwise.
Understanding and identifying p-series is vital, as they frequently appear or are used as benchmark tests. For instance, they are frequently used in the Comparison Test to assist in evaluating the convergence of more intricate series. They provide a clear threshold for convergence predicated simply on the value of \( p \), which is evident in the earlier analysis with series 'a', where a comparison to a known convergent p-series (\( p = 2 \)) simplified concluding its convergence. Consequently, the p-series is not only a building block for compatibility tests but also empowers intuition regarding series behavior.
Other exercises in this chapter
Problem 1
For those sequences that converge, find the limit \(\lim _{n \rightarrow \infty} a_{n}\) a. \(a_{n}=\frac{n^{2}+1}{n^{3}+1} .\) b. \(a_{n}=\frac{3 n+1}{n+2}\).
View solution Problem 2
Find the sum for each of the series: a. \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 3}{4^{n}}\) b. \(\sum_{n=2}^{\infty} \frac{2}{5^{n}}\). c. \(\sum_{n=0}^{\infty}\le
View solution Problem 4
Do the following: a. Compute: \(\lim _{n \rightarrow \infty} n \ln \left(1-\frac{3}{n}\right)\). b. Use L'Hopital's Rule to evaluate \(L=\lim _{x \rightarrow \i
View solution Problem 5
Consider the sum \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}\) a. Use an appropriate convergence test to show that this series converges. b. Verify that $$ \sum_
View solution