Problem 19 - Essential Calculus Early Transcendentals

Question

Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \sqrt[n]{2}$$

Step-by-Step Solution

Answer

The series is divergent since the terms do not approach zero.

Step 1: Understand the Series

The series given is \( \sum_{n=1}^{\infty} \sqrt[n]{2} \), which is equivalent to \( \sum_{n=1}^{\infty} 2^{1/n} \). We need to determine if this series is convergent or divergent.

Step 2: Analyze the General Term

The general term of the series is \( a_n = 2^{1/n} \). As \( n \to \infty \), \( 2^{1/n} \to 1 \). Since \( a_n \) does not approach 0, the series does not meet the necessary condition for convergence.

Step 3: Conclusion on Convergence

Because the terms of the series \( a_n = 2^{1/n} \) do not approach 0 as \( n \) approaches infinity, the series \( \sum_{n=1}^{\infty} 2^{1/n} \) is divergent.

Key Concepts

General Term Analysis

In an infinite series, the general term is a crucial element that provides insight into the behavior of the series. When we are given a series, such as \( \sum_{n=1}^{\infty} \sqrt[n]{2} \), it's important to express it in terms of its general term for easier understanding and analysis. In this case, the general term of the series is represented as \( a_n = 2^{1/n} \).
Analyzing the general term involves examining its behavior as \( n \) increases. A major point of interest is what happens to \( a_n \) as \( n \to \infty \). For our series, as \( n \) becomes larger, \( 2^{1/n} \) approaches 1. This pattern or behavior will be fundamental in determining whether the series converges or diverges. Understanding how the general term evolves sets the stage for the next steps, such as verifying conditions like the necessary condition for convergence.

Necessary Condition for Convergence

A very fundamental rule in series analysis is the necessary condition for the convergence of an infinite series. For a series \( \sum_{n=1}^{\infty} a_n \) to converge, the terms \( a_n \) must approach 0 as \( n \to \infty \). This is a necessary but not sufficient condition, which means that while it may not guarantee convergence on its own, failure to meet this condition certainly implies divergence.
In the series \( \sum_{n=1}^{\infty} 2^{1/n} \), examining the general term \( a_n = 2^{1/n} \) shows us that \( a_n \to 1 \) as \( n \to \infty \). Since the necessary condition of \( a_n \to 0 \) is not satisfied, we can immediately conclude that this series cannot be convergent. If a series fails to meet this necessary condition, it saves you from needing to perform more complicated tests for convergence.

Infinite Series

Infinite series can initially seem quite complex, but they represent just an extension of finite series, where the series continues indefinitely. Understanding that we're summing an infinite number of terms helps in recognizing patterns or behaviors that differentiate convergent and divergent series.
The series \( \sum_{n=1}^{\infty} 2^{1/n} \) is one such series, where we’re adding an infinite number of terms. The key question to ask is: does this running total add up to a specific number, or does it grow without bound? If the sum approaches a specific value as you add more terms, the series is convergent.
However, in this particular series, due to the pattern of the general term that doesn't tend to zero, we find it grows unbounded, confirming its divergence. Being able to classify an infinite series quickly — by checking such conditions as the behavior of its general terms — forms a strong basis for further study and application in various mathematical and scientific fields.