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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.

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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.
  • Divergent series do not settle at a single value; they grow infinitely.
  • Recognizing divergence helps in determining the behavior of similar series.
A divergent series contrasts with a convergent series, which does approach a finite bound.
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.
  • 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.
It serves as a benchmark in calculus and helps with comparing other series using test methods like the Comparison Test.
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:
  • \( \frac{1}{n+1} < \frac{1}{n} \)
  • Because \( \sum \frac{1}{n} \) diverges, by the Comparison Test, \( \sum \frac{1}{n+1} \) must also diverge.
This logic extends to the series \( \frac{1}{2} \sum \frac{1}{n+1} \), since multiplying by a constant does not change the divergence of a series. When a constant factor multiplies a divergent series, it simply scales the divergence, but doesn't prevent it from diverging.

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Most popular questions from this chapter

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges. Let \(a_{n}=\frac{\pi^{\ln n}}{(\ln n)^{n}} .\) Show that \(\frac{a_{2 n}}{a_{n}} \rightarrow 0\) as \(n \rightarrow \infty\).

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.Let \(a_{n}^{+}=a_{n}\) if \(a_{n} \geq 0\) and \(a_{n}^{-}=-a_{n}\) if \(a_{n}<0 .\) (Also, \(a_{n}^{+}=0\) if \(a_{n}<0\) and \(a_{n}^{-}=0\) if \(\left.a_{n} \geq 0 .\right)\) If \(\sum_{n=1}^{\infty} a_{n}\) converges conditionally but not absolutely, then neither \(\sum_{n=1}^{\infty} a_{n}^{+}\) nor \(\sum_{n=1}^{\infty} a_{n}^{-}\) converge.

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit. $$ a_{n}=\frac{2^{n+1}}{5^{n}} $$

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, or state if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{2^{n^{2}}}{n^{n} n !} $$

Does there exist a number \(p\) such that \(\sum_{n=1}^{\infty} \frac{2^{n}}{n^{p}}\) converges?

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