Problem 34
Question
Use any test to determine whether the series is absolutely convergent, conditionally convergent, or divergent. \( \displaystyle \sum_{n = 1}^{\infty} \frac {n5^{2n}}{10^{n+1}} \)
Step-by-Step Solution
Verified Answer
The series is divergent.
1Step 1: Write the Series in Standard Form
The series given is \( \sum_{n = 1}^{\infty} \frac{n5^{2n}}{10^{n+1}} \). We can simplify this to \( \sum_{n = 1}^{\infty} \frac{n5^{2n}}{10^n \cdot 10} = \frac{1}{10} \sum_{n = 1}^{\infty} \frac{n5^{2n}}{10^n} \). This is now clearer to analyze.
2Step 2: Analyze the General Term
Rewrite the general term of the series as \( \frac{n5^{2n}}{10^n} = n \left(\frac{25}{10}\right)^n = n \left(\frac{5}{2}\right)^n \). Notice \( \left(\frac{5}{2}\right) \) is greater than 1.
3Step 3: Apply the Ratio Test
For the ratio test, find \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Here, \( a_n = n \left(\frac{5}{2}\right)^n \). We calculate \( \lim_{n \to \infty} \frac{(n+1) \left(\frac{5}{2}\right)^{n+1}}{n \left(\frac{5}{2}\right)^n} = \frac{5}{2} \cdot \lim_{n \to \infty} \frac{n+1}{n} = \frac{5}{2} \) as the common factors cancel, and the limit simplifies to \( \frac{5}{2} \).
4Step 4: Interpret the Ratio Test Result
Since \( \frac{5}{2} > 1 \), the ratio test indicates that the series \( \sum_{n = 1}^{\infty} \frac{n5^{2n}}{10^{n}} \) diverges. Therefore, it is neither absolutely convergent nor conditionally convergent.
Key Concepts
Ratio TestAbsolute ConvergenceConditional ConvergenceDivergent Series
Ratio Test
The Ratio Test is a handy tool used to determine the convergence or divergence of infinite series. It works particularly well on series where each term includes variable exponents or factorials.
To apply the Ratio Test, you first take the absolute value of the ratio of consecutive terms. For a series \( \sum a_n \), you'll want to check:
To apply the Ratio Test, you first take the absolute value of the ratio of consecutive terms. For a series \( \sum a_n \), you'll want to check:
- \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)
Absolute Convergence
Absolute Convergence is when a series remains convergent even when we take the absolute value of each term. This concept is crucial because it guarantees a strong form of convergence where rearranging terms will not change the sum of the series.
To check for absolute convergence, you look at the series formed by taking the absolute value of each term. If \( \sum |a_n| \) converges, then the series \( \sum a_n \) also converges absolutely.
In our exercise with \( \sum_{n = 1}^{\infty} \frac{n5^{2n}}{10^{n+1}} \), considering the result of the Ratio Test gave a value greater than 1, indicating not only divergence but it also signifies that the series does not have absolute convergence.
To check for absolute convergence, you look at the series formed by taking the absolute value of each term. If \( \sum |a_n| \) converges, then the series \( \sum a_n \) also converges absolutely.
In our exercise with \( \sum_{n = 1}^{\infty} \frac{n5^{2n}}{10^{n+1}} \), considering the result of the Ratio Test gave a value greater than 1, indicating not only divergence but it also signifies that the series does not have absolute convergence.
Conditional Convergence
Conditional Convergence occurs when a series converges, but it fails to converge absolutely. This means the series \( \sum a_n \) converges, but \( \sum |a_n| \) does not.
Most famously seen in alternating series, conditional convergence suggests that some terms' positive and negative values are balancing to achieve convergence.
In this particular example, the result from the original step-by-step solution shows the series is divergent, indicating it does not achieve conditional convergence either. As a series must converge in some form to be conditionally convergent, the findings explicitly suggest the absence of both absolute and conditional convergence.
Most famously seen in alternating series, conditional convergence suggests that some terms' positive and negative values are balancing to achieve convergence.
In this particular example, the result from the original step-by-step solution shows the series is divergent, indicating it does not achieve conditional convergence either. As a series must converge in some form to be conditionally convergent, the findings explicitly suggest the absence of both absolute and conditional convergence.
Divergent Series
A Divergent Series is one that does not settle to a finite limit. In simpler terms, the sum keeps growing larger or does not settle to a constant as more terms are added.
In our specific exercise, using the Ratio Test revealed a limit of \( \frac{5}{2} \), greater than 1. This signifies that the terms of the series do not approach zero fast enough for the entire series to have a sum that stabilizes.
Understanding divergence in this context highlights the importance of the various tests, such as the Ratio Test, in determining the behavior of series. When a series is divergent, it provides the important insight that mathematicians or scientists cannot rely on it for finite results, a crucial realization for further mathematical analysis.
In our specific exercise, using the Ratio Test revealed a limit of \( \frac{5}{2} \), greater than 1. This signifies that the terms of the series do not approach zero fast enough for the entire series to have a sum that stabilizes.
Understanding divergence in this context highlights the importance of the various tests, such as the Ratio Test, in determining the behavior of series. When a series is divergent, it provides the important insight that mathematicians or scientists cannot rely on it for finite results, a crucial realization for further mathematical analysis.
Other exercises in this chapter
Problem 34
Graph the first several partial sums \( s_n (x) \) of the series \( \sum_{n = 0}^{\infty} x^n, \) together with the sum function \( f(x) = 1/(1 - x), \) on a co
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Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n + n \cos^2n} \)
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For what value of \( p \) is each series convergent? \( \displaystyle \sum_{n = 2}^{\infty} ( - 1)^{n-1} \frac {(\ln n)^P}{n} \)
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Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error. \( \displaystyle \sum_{n = 1}^{\infty} \frac {e^{1/n}}{n^4} \)
View solution