Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 7–12, show that each series converges absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^{2}}{e^{n}} $$

Step-by-Step Solution

Answer

The series converges absolutely since the series of absolute values converges.

Step 1: Understand Absolute Convergence

A series \( \sum_{n=1}^{\infty} a_n \) is said to converge absolutely if the series \( \sum_{n=1}^{\infty} |a_n| \) converges. For the given series, the terms are \( a_n = (-1)^{n+1} \frac{n^2}{e^n} \). We must first consider the series \( \sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{n^2}{e^n} \right| = \sum_{n=1}^{\infty} \frac{n^2}{e^n} \).

Step 2: Apply the Ratio Test

To determine if \( \sum_{n=1}^{\infty} \frac{n^2}{e^n} \) converges, we can use the Ratio Test, which states that if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), the series converges absolutely. For our series, \( a_n = \frac{n^2}{e^n} \) and \( a_{n+1} = \frac{(n+1)^2}{e^{n+1}} \).

Step 3: Calculate the Ratio

Calculate the ratio: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)^2 e^n}{n^2 e^{n+1}} \right| = \frac{(n+1)^2}{n^2 e} \]This simplifies to \[ \frac{(n+1)^2}{n^2 e} = \frac{n^2 + 2n + 1}{n^2 e} = \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{e} \].

Step 4: Take the Limit

To apply the Ratio Test, take the limit of the ratio as \( n \to \infty \):\[ \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{e} = \frac{1}{e} \]Since \( \frac{1}{e} < 1 \), the Ratio Test confirms that \( \sum_{n=1}^{\infty} \frac{n^2}{e^n} \) converges.

Step 5: Conclude Absolute Convergence

Since the series \( \sum_{n=1}^{\infty} \frac{n^2}{e^n} \) converges, the original series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n^2}{e^n} \) converges absolutely by the definition of absolute convergence.

Key Concepts

Ratio Test

The Ratio Test is a powerful tool used to determine the absolute convergence of a series. It applies nicely to series where the terms become notably smaller quickly. Here's how the Ratio Test works:

• Take the expression for the term of the series, say, \( a_n \).
• Find the next term in the sequence, \( a_{n+1} \).
• Calculate the ratio: \( \left| \frac{a_{n+1}}{a_n} \right| \).
• Determine the limit of this ratio as \( n \) approaches infinity: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

If this limit is less than 1, the series converges absolutely. If the limit is greater than 1, the series diverges. And if it equals 1, the test is inconclusive. In the given problem, we've applied this test to \( \sum_{n=1}^{\infty} \frac{n^2}{e^n} \), and found that the limit is \( \frac{1}{e} \), which is less than 1, ensuring convergence.

Series Convergence

Series convergence is the idea that the sum of an infinite series can approach a finite number. This concept is crucial in calculus as it helps us understand how series behave over long terms. There are several ways to test for convergence:

• Geometric Series Test: Used when each term is a constant multiplied by the previous term.
• p-Series Test: Helps when the series is of the form \( \sum \frac{1}{n^p} \).
• Comparison Test: Compares a series to a known convergent or divergent series.
• Ratio Test: Often used for series with factorials or exponential terms, as explained previously.

In the problem discussed, using the Ratio Test confirmed that the series \( \sum_{n=1}^{\infty} \frac{n^2}{e^n} \) converges, meaning that the sum of all its infinite terms reaches a finite value.

Calculus Series

Infinite series are a cornerstone of calculus. They allow us to express functions as sums of simpler terms, making complex calculations more manageable. Calculus series get into integrating and differentiating function representations in series form.
There are various series types:

• Power Series: These express functions as sums of power terms \( a_n x^n \).
• Taylor Series: A type of power series that represents functions using derivatives at a specific point.
• Fourier Series: Allows periodic functions to be expressed as sums of sines and cosines.

The calculus series in the problem involves exponential components, \( e^n \), which makes the Ratio Test a preferred method for assessing convergence due to its efficiency in dealing with exponentials. Understanding series helps in approximating functions and solving complex equations that are central in calculus.