Problem 17 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{1}{2 \sqrt{n}+\sqrt[3]{n}}\end{equation}

Step-by-Step Solution

Answer

The series diverges by the Limit Comparison Test.

Step 1: Analyze the Series

We start by considering the series \( \sum_{n=1}^{\infty} \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \). We need to determine if it converges or diverges. Notice the terms are positive, so we may use comparison tests.

Step 2: Simplify Expression

Consider the term \( \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \). For large \( n \), \( 2\sqrt{n} \) will dominate over \( \sqrt[3]{n} \). We can simplify by approximating \( 2\sqrt{n} + \sqrt[3]{n} \approx 2\sqrt{n} \). Thus, the term \( \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \approx \frac{1}{2\sqrt{n}} \).

Step 3: Comparison with a Known Series

The series \( \sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) is a \( p \)-series where \( p = \frac{1}{2} \lt 1 \). We know that \( p \)-series diverge if \( p \leq 1 \). Hence, \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) diverges.

Step 4: Apply the Limit Comparison Test

To make our conclusion rigorous, use the Limit Comparison Test: compare \( a_n = \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \) with \( b_n = \frac{1}{\sqrt{n}} \). Compute \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \cdot \sqrt{n} = \lim_{n \to \infty} \frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{\sqrt[6]{n}}} = \frac{1}{2} \).

Step 5: Conclusion on Convergence/Divergence

Since the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{1}{2} \) is a positive and finite number, and \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) diverges, the Limit Comparison Test tells us that \( \sum_{n=1}^{\infty} \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \) also diverges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is an essential tool in determining the convergence or divergence of a series. It is particularly useful when we encounter complex series that don't straightforwardly fit into other easier convergence tests. The basic idea is to compare the complex series to a simpler one, where the behavior (convergence or divergence) is already known.
To apply the Limit Comparison Test, we select two series: the one we are interested in, denoted as \( a_n \), and a known series \( b_n \) that behaves similarly for large \( n \). We then evaluate the limit:

• \( \lim_{n \to \infty} \frac{a_n}{b_n} \)

If this limit evaluates to a positive real number, it implies that both series, \( \sum a_n \) and \( \sum b_n \), either converge together or diverge together. In the original problem, the complicated series term \( a_n = \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \) was compared with the well-known \( b_n = \frac{1}{\sqrt{n}} \). The limit calculated was \( \frac{1}{2} \), which is positive and finite, indicating that both series share the same convergence behavior.

P-series

P-series plays a crucial role in analyzing series convergence because of its straightforward criteria. A p-series is of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The convergence of a p-series depends on the value of \( p \):

• Converges if \( p > 1 \)
• Diverges if \( p \leq 1 \)

In the original exercise, by simplifying \( \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \) to \( \frac{1}{2\sqrt{n}} \), we recognized it as part of a p-series with \( p = \frac{1}{2} \). Since \( \frac{1}{2} < 1 \), this series diverges. Understanding p-series helps frame complex series in simpler forms, making it easier to apply other convergence tests like the Limit Comparison Test.

Divergence of Series

The concept of divergence is as significant as convergence in the study of series. When a series diverges, it means that the sum of its terms does not approach a finite limit. There are several ways to determine whether a series diverges. Apart from the p-series test, we can also use comparison tests to establish divergence.
In the context of the original exercise, the series \( \sum_{n=1}^{\infty} \frac{1}{2\sqrt{n} + \sqrt[3]{n}} \) diverged. Upon using the Limit Comparison Test, it was compared with the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \), which is known to diverge. Since the original series is similarly divergent, the comparison test provided a conclusive answer.
It is vital always to check the foundational properties of the series before proceeding with advanced tests. Understanding divergence forms the basis for deciding the method of testing a series, and this forms the groundwork for deeper analysis in calculus.