Problem 63 - Thomas Calculus

Question

In Exercises \(57 - 82 ,\) use any method to determine whether the series converges or diverges. Give reasons for your answer. $$ \sum _ { n = 1 } ^ { \infty } n ^ { - 2 / \sqrt { 5 } } $$

Step-by-Step Solution

Answer

The series diverges by the p-series test.

Step 1: Identify the Series Type

The series in question is \( \sum _ { n = 1 } ^ { \infty } n ^ { - 2 / \sqrt { 5 } } \). This series is a \emph{p-series} with the exponent \( p \) given by \( p = \frac{2}{\sqrt{5}} \). The form of a \emph{p-series} is \( \sum _ { n = 1 } ^ { \infty } n^{-p} \).

Step 2: Determine the Convergence of a p-Series

A \emph{p-series} \( \sum _ { n = 1 } ^ { \infty } n^{-p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \). In this problem, we need to determine if \( \frac{2}{\sqrt{5}} > 1 \).

Step 3: Evaluate the Exponent

To evaluate \( \frac{2}{\sqrt{5}} \), calculate its numerical value. \( \sqrt{5} \approx 2.236 \), thus, \( \frac{2}{\sqrt{5}} \approx 0.8944 \).

Step 4: Conclude the Convergence or Divergence

Since \( \frac{2}{\sqrt{5}} \approx 0.8944 \) is less than 1, the original series \( \sum _ { n = 1 } ^ { \infty } n ^ { - 2 / \sqrt { 5 } } \) is a \emph{p-series} with \( p < 1 \). Therefore, the series diverges by the \emph{p-series} test.

Key Concepts

Convergence

In mathematics, **convergence** is a term used to describe a series that approaches a specific number or value as the number of terms increases. To test for convergence, we evaluate whether the sum of a series yields a finite result. With **p-series**, convergence depends on the value of the exponent, denoted as \( p \). For a p-series written as \[\sum _ { n = 1 } ^ { \infty } n^{-p},\]the series is said to converge if \( p > 1 \). In this exercise, we look at whether \( \frac{2}{\sqrt{5}} \) is greater than 1. Calculating this gives \( 0.8944 \), which does not satisfy the convergence condition \( p > 1 \). Therefore, the series does not converge, meaning the sum does not settle to a finite number.

Divergence

**Divergence** implies that a series does not approach a finite limit, and its sum tends towards infinity, or it does not settle around any particular number as its limit. For **p-series**, divergence occurs when the value of \( p \) is less than or equal to 1. The series given in the exercise, \[\sum _ { n = 1 } ^ { \infty } n ^ { - 2 / \sqrt { 5 } },\]has an exponent \( p = \frac{2}{\sqrt{5}} \) which approximately equals \( 0.8944 \). Since \( p \approx 0.8944 \leq 1 \), by the **p-series test**, this series will diverge. In simple terms, it won't settle or converge to a particular number.

Series Test

The **series test** used here is the **p-series test**, specifically designed for series of the form \[\sum _ { n = 1 } ^ { \infty } n^{-p}.\]This test determines whether a series converges or diverges based on the value of \( p \), where:

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

For our exercise, we determine that \( \frac{2}{\sqrt{5}} \approx 0.8944 \). Applying the **p-series test**, we note that since \( 0.8944 \leq 1 \), the series satisfies the divergence condition. Thus, the original series diverges. This makes the **p-series test** a simple yet powerful tool to discern convergence or divergence in such series types.