Problem 26 - Calculus: Early Transcendentals

Question

Use the Root Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 1}^{\infty} \frac {( - 2)^n}{n^n} \)

Step-by-Step Solution

Answer

The series is convergent.

Step 1: Write the Terms of the Series

The given series is \( \sum_{n=1}^{\infty} \frac{(-2)^n}{n^n} \). Each term in this series can be represented as \( a_n = \frac{(-2)^n}{n^n} \).

Step 2: Apply the Root Test

The Root Test requires us to compute \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \). Substitute \( a_n = \frac{(-2)^n}{n^n} \) to get \( a_n = \frac{|(-2)^n|}{n^n} = \frac{2^n}{n^n} \). Now, we need \( \lim_{n \to \infty} \sqrt[n]{\frac{2^n}{n^n}} \).

Step 3: Simplify the Expression Inside the Root

Simplifying, we have \( \sqrt[n]{\frac{2^n}{n^n}} = \frac{\sqrt[n]{2^n}}{\sqrt[n]{n^n}} = \frac{2}{n} \).

Step 4: Compute the Limit

Now compute the limit: \( \lim_{n \to \infty} \frac{2}{n} = 0 \). As \( n \) approaches infinity, \( \frac{2}{n} \) approaches 0.

Step 5: Conclude Based on Root Test Results

The Root Test states that if \( \lim_{n \to \infty} \sqrt[n]{|a_n|} < 1 \), then the series \( \sum a_n \) is absolutely convergent. Since \( \lim_{n \to \infty} \frac{2}{n} = 0 < 1 \), the series is convergent.

Key Concepts

Convergent Series

A convergent series is one that approaches a specific limit as the number of terms increases indefinitely. In simpler terms, it "settles down" to a particular number, rather than heading towards infinity. When working with series, it's important to establish whether they are convergent because it tells us about its behavior and sum in the long term. In the case of the given series, using the Root Test helps in determining its convergence.
One key property of convergent series is that the terms keep getting smaller, ultimately nearing zero. For the series \( \sum_{n=1}^{\infty} \frac{(-2)^n}{n^n} \), the conditions of the Root Test confirm that this series converges. This insight means that as you add more terms, their contributions become negligible, leading the series to stabilize towards a particular value.

Divergent Series

Unlike convergent series, a divergent series does not approach a particular limit. Instead, its terms continue to grow indefinitely or bounce around without settling down as you sum them over time. This unpredictable behavior can result in the series heading towards various values without showing any sign of convergence.
Divergence is crucial to identify since it indicates that the series doesn't have a finite sum. In essence, if we consider the opposite of the behavior we observed in the original exercise's series, where the limit was less than one, a divergent series would not meet such a criterion. Using the Root Test in our case highlighted that the given series was not divergent because the root of the absolute term was less than one.

Limit Calculation

Limit calculation is a fundamental concept used to determine the behavior of series and functions as they approach a specific point, typically infinity. In calculating the limit, you assess what happens to the series as it continues indefinitely. This process is critical for applying tests such as the Root Test.
In our exercise, we calculated the limit by evaluating \( \lim_{n \to \infty} \frac{2}{n} \). This calculation involved simplifying the expression inside the root to see its behavior as \( n \) becomes very large. The result was zero, indicating that the terms decrease, affirming convergence under the Root Test's criterion. Understanding this process helps in predicting the behavior of more complex series effectively.

Absolute Convergence

Absolute convergence goes a step further in analyzing series, implying that even after taking the absolute value of all terms, the series still converges. It is a stronger form of convergence, reassuring us that fluctuations, including those caused by negative terms, do not affect the stability of the series.
For the series \( \sum_{n=1}^{\infty} \frac{(-2)^n}{n^n} \), absolute convergence was verified using the Root Test. The test confirmed that \( \lim_{n \to \infty} \sqrt[n]{|a_n|} < 1 \), securing its absolute convergence. This means that even if negative or alternating terms are present, their absolute values form a series that converges. Evaluating absolute convergence is significant for series with terms alternating in sign, ensuring overall stability and finite sums.