Chapter 4: Problem 3
Use the comparison test to determine whether the following series converge. $$ \sum_{n=1}^{\infty} \frac{1}{2(n+1)} $$
Short Answer
Expert verified
The series diverges.
Step by step solution
01
Identify the Series
The given series is \( \sum_{n=1}^{\infty} \frac{1}{2(n+1)} \). This can be simplified to \( \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n+1} \), highlighting the factor of \( \frac{1}{2} \) outside of the summation.
02
Choose a Comparison Series
We compare \( \frac{1}{n+1} \) with the harmonic series \( \frac{1}{n} \), a known divergent series. Because \( \frac{1}{n+1} \approx \frac{1}{n} \) for large \( n \), it serves as a good candidate for comparison.
03
Apply the Comparison Test
By the Comparison Test, since \( \frac{1}{n+1} < \frac{1}{n} \), if \( \sum \frac{1}{n} \) diverges, \( \sum \frac{1}{n+1} \) also diverges because it compares to a known larger divergent series.
04
Conclusion on Convergence
Since \( \frac{1}{n+1} \) diverges and the given series is \( \frac{1}{2} \sum \frac{1}{n+1} \), multiplying by a constant factor \( \frac{1}{2} \) does not affect divergence. Therefore, the original series \( \sum_{n=1}^{\infty} \frac{1}{2(n+1)} \) also diverges.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Divergent Series
A divergent series is a sequence of numbers that does not approach a finite limit as more terms are added. This means, no matter how many terms you include, the series does not sum up to a fixed number. Instead, it grows indefinitely. A classic example of a divergent series is the harmonic series:
\[\sum_{n=1}^{\infty} \frac{1}{n}\]In this series, each term is the reciprocal of a positive integer. As you keep adding these reciprocals, the sum grows beyond any measurable limit. Understanding the concept of divergence is crucial when analyzing series using the Comparison Test.
\[\sum_{n=1}^{\infty} \frac{1}{n}\]In this series, each term is the reciprocal of a positive integer. As you keep adding these reciprocals, the sum grows beyond any measurable limit. Understanding the concept of divergence is crucial when analyzing series using the Comparison Test.
- Divergent series do not settle at a single value; they grow infinitely.
- Recognizing divergence helps in determining the behavior of similar series.
Harmonic Series
The harmonic series is an important example of a series that diverges. It takes the form
\[\sum_{n=1}^{\infty} \frac{1}{n}\]Each term in this series is the reciprocal of a natural number. Although harmonic series grow very slowly compared to geometric progressions, they still eventually diverge. This means their sum increases beyond limits as we add more terms.
\[\sum_{n=1}^{\infty} \frac{1}{n}\]Each term in this series is the reciprocal of a natural number. Although harmonic series grow very slowly compared to geometric progressions, they still eventually diverge. This means their sum increases beyond limits as we add more terms.
- Slow growth rate does not mean convergence, as seen with the harmonic series.
- Despite each term decreasing towards zero, the series itself does not converge.
Convergence of Series
The convergence of a series is when the sum of its terms approaches a specific value as more terms are added. In the context of determining convergence, the Comparison Test is a valuable tool. It requires comparing the series in question with another series whose behavior is known, such as the harmonic series.
By comparing the series \( \sum \frac{1}{n+1} \) with the harmonic series \( \sum \frac{1}{n} \), we use the fact that:
By comparing the series \( \sum \frac{1}{n+1} \) with the harmonic series \( \sum \frac{1}{n} \), we use the fact that:
- \( \frac{1}{n+1} < \frac{1}{n} \)
- Because \( \sum \frac{1}{n} \) diverges, by the Comparison Test, \( \sum \frac{1}{n+1} \) must also diverge.