Chapter 9: Problem 10
Show that each series converges absolutely. \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^{2}}{e^{n}}\)
Short Answer
Expert verified
The series converges absolutely by the Ratio Test.
Step by step solution
01
Understand Absolute Convergence
A series \(\sum a_n\) converges absolutely if the series of absolute values \(\sum |a_n|\) converges. For the given series, \(a_n = (-1)^{n+1} \frac{n^2}{e^n}\), the series of absolute values is \(\sum_{n=1}^{\infty} \frac{n^2}{e^n}\). We need to check the convergence of \(\sum_{n=1}^{\infty} \frac{n^2}{e^n}\).
02
Use the Ratio Test for Convergence
The Ratio Test states that a series \(\sum a_n\) converges absolutely if the limit \(L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\). Apply the Ratio Test to \(\sum \frac{n^2}{e^n}\):\[L = \lim_{n\to\infty} \frac{(n+1)^2/e^{n+1}}{n^2/e^n} = \lim_{n\to\infty} \frac{(n+1)^2}{n^2}\cdot\frac{1}{e} = \lim_{n\to\infty} \frac{n^2 + 2n + 1}{n^2}\cdot\frac{1}{e} = \lim_{n\to\infty} \left(1 + \frac{2}{n} + \frac{1}{n^2}\right) \cdot \frac{1}{e}.\]As \(n\to\infty\), \(\frac{2}{n} \to 0\) and \(\frac{1}{n^2} \to 0\), so \(L = \frac{1}{e} < 1\).
03
Conclude Absolute Convergence
Since \(L < 1\), the Ratio Test confirms that \(\sum_{n=1}^{\infty} \frac{n^2}{e^n}\) converges. Therefore, the original series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{e^n}\) converges absolutely.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ratio Test
The Ratio Test is a powerful tool used to determine whether a series converges absolutely. It comes in handy when you have complex expressions that are changing in magnitude. Here's how it works: the test considers the ratio of consecutive terms in the series.
- You calculate the ratio \( |a_{n+1} / a_n| \) of each term.
- Find the limit \( L = \lim_{n \to \infty} |a_{n+1} / a_n| \).
- If this limit \( L \) is less than 1, the series converges absolutely.
- If \( L \) is greater than 1, the series diverges.
- If \( L \) is equal to 1, the test is inconclusive.
Convergent Series
A convergent series is a series whose sequence of partial sums approaches a specific value. When working with convergent series, it is important to understand that:
- If the series converges, the terms of the series get smaller, approaching zero as more terms are added.
- Each step of summing comes closer to a particular limit.
- No matter how far you go in the series, the sum does not jump away from this limit.
Power Series
A power series is another interesting concept related to convergence. A power series takes the form \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where \( c_n \) are coefficients and \( (x-a) \) is raised to the power of \( n \). Power series are encountered frequently within calculus and analysis, offering profound insights and applications, including representing functions analytically.
- The general form involves a center \( a \) and varies with \( x \).
- Convergence of a power series depends on the value of \( x \) and requires determining the radius and interval of convergence.
- Power series can converge absolutely or conditionally within its interval of convergence.