Problem 32
Question
Use any test to determine whether the series is absolutely convergent, conditionally convergent, or divergent. \( \displaystyle \sum_{n = 1}^{\infty} \left( \frac {1 - n}{2 + 3n} \right) ^n \)
Step-by-Step Solution
Verified Answer
The series is absolutely convergent.
1Step 1: Recognize the Series Type
We are given the series \( \sum_{n=1}^{\infty} \left( \frac{1-n}{2+3n} \right)^n \). This is a series with terms that are raised to their corresponding index \( n \). This hints that the root test might be appropriate to determine the series' convergence.
2Step 2: Apply the Root Test
The root test states that for a series \( \sum a_n \), if \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = L \), then if \( L < 1 \), the series converges absolutely; if \( L > 1 \), the series diverges; and if \( L = 1 \), the test is inconclusive. We set \( a_n = \left( \frac{1-n}{2+3n} \right)^n \) and find \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \).
3Step 3: Calculate \( |a_n| \)
For \( |a_n| = \left| \left( \frac{1-n}{2+3n} \right)^n \right| \), we have \( |a_n| = \left( \frac{n-1}{2+3n} \right)^n \) because \( n > 1 \) and hence \( 1-n < 0 \).
4Step 4: Simplify \( \sqrt[n]{|a_n|} \)
Compute \( \lim_{n \to \infty} \sqrt[n]{\left( \frac{n-1}{2+3n} \right)^n} = \lim_{n \to \infty} \frac{n-1}{2+3n} \).
5Step 5: Evaluate the Limit
Simplify the limit: \( \lim_{n \to \infty} \frac{n-1}{2+3n} = \lim_{n \to \infty} \frac{1 - \frac{1}{n}}{\frac{2}{n} + 3} = \frac{1 - 0}{0 + 3} = \frac{1}{3} \).
6Step 6: Conclude using Root Test
Since \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = \frac{1}{3} \), which is less than 1, the series \( \sum_{n=1}^{\infty} \left( \frac{1-n}{2+3n} \right)^n \) is absolutely convergent.
Key Concepts
Absolutely ConvergentConditionally ConvergentRoot TestSeries Convergence
Absolutely Convergent
A series is considered absolutely convergent if the series of absolute values of its terms also converges. In simpler terms, if you take the absolute value of each term in a series and the resulting series still converges, then the original series is absolutely convergent.
When a series is absolutely convergent, it implies a stronger form of convergence than conditional convergence. Absolutely convergent series have the property of unconditional rearrangement. This means that you can permute (rearrange) the terms of an absolutely convergent series in any way, and it will still converge to the same sum.
For instance, the series \[ \sum_{n=1}^{\infty} \left( \frac{1-n}{2+3n} \right)^n \] is absolutely convergent because, using the root test, we found that the series of its absolute values converges.
When a series is absolutely convergent, it implies a stronger form of convergence than conditional convergence. Absolutely convergent series have the property of unconditional rearrangement. This means that you can permute (rearrange) the terms of an absolutely convergent series in any way, and it will still converge to the same sum.
For instance, the series \[ \sum_{n=1}^{\infty} \left( \frac{1-n}{2+3n} \right)^n \] is absolutely convergent because, using the root test, we found that the series of its absolute values converges.
Conditionally Convergent
A series is conditionally convergent if it converges, but it does not converge absolutely. Essentially, this means that while the series converges, the series formed by taking the absolute values of its terms diverges.
The classic example of a conditionally convergent series is the alternating harmonic series. Conditionally convergent series have different properties than absolutely convergent series. One key difference is that the sum of a conditionally convergent series can change based on how its terms are rearranged.
In the original exercise, after applying the root test, we determined the series converges absolutely, so it cannot be conditionally convergent.
The classic example of a conditionally convergent series is the alternating harmonic series. Conditionally convergent series have different properties than absolutely convergent series. One key difference is that the sum of a conditionally convergent series can change based on how its terms are rearranged.
In the original exercise, after applying the root test, we determined the series converges absolutely, so it cannot be conditionally convergent.
Root Test
The root test, also known as the Cauchy root test, is a method used to determine the convergence of a series. Use this test when the terms of the series are raised to powers involving the index number, which is similar to our given series: \( \sum_{n=1}^{\infty} \left( \frac{1-n}{2+3n} \right)^n \).
- If \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = L < 1 \), the series converges absolutely.
- If \( L > 1 \), the series diverges.
- If \( L = 1 \), the test is inconclusive and other convergence tests might be more useful.
Series Convergence
Series convergence is a fundamental concept in calculus and analysis. It involves generating a sum from an infinite sequence of terms. A series can converge to a specific value or diverge if it does not settle on a value as more terms are added.
There are various tests to determine whether a series converges, including the root test, ratio test, and comparison test, each suited to different series structures. One must choose an appropriate test based on the form of the series in question.
The exercise at hand utilized the root test to discover the convergence properties of the series \( \sum_{n=1}^{\infty} \left( \frac{1-n}{2+3n} \right)^n \). This series was shown to converge absolutely, indicating that the sum of its absolute values also converges.
There are various tests to determine whether a series converges, including the root test, ratio test, and comparison test, each suited to different series structures. One must choose an appropriate test based on the form of the series in question.
The exercise at hand utilized the root test to discover the convergence properties of the series \( \sum_{n=1}^{\infty} \left( \frac{1-n}{2+3n} \right)^n \). This series was shown to converge absolutely, indicating that the sum of its absolute values also converges.
Other exercises in this chapter
Problem 32
Let \( p \) and \( q \) be real numbers with \( p
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Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \frac {(n!)^n}{n^{4n}} \)
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For what value of \( p \) is each series convergent? \( \displaystyle \sum_{n = 1}^{\infty} \frac {( - 1)^{n-1}}{n^p} \)
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Determine whether the series converges or diverges. \( \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n^{1 + 1/n}} \)
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