Problem 51 - 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}{n \sqrt[n]{n}}\end{equation}

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}}\) converges.

Step 1: Identify the Series Type

He have the series \(\sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}}\). We note that it resembles a form of the p-series, but with an additional root in the term. This suggests the use of a comparison or root test might be suitable.

Step 2: Apply the Root Test

The root test involves the limit \( L = \lim_{n \to \infty} \sqrt[n]{a_n} \). For our series, \(a_n = \frac{1}{n \sqrt[n]{n}}\). Thus, \(\sqrt[n]{a_n} = \frac{1}{n^{1/n}}\). We find the limit: \[L = \lim_{n \to \infty} \frac{1}{n^{1/n}} = \frac{1}{1} = 1.\]Since \(L = 1\), the root test is inconclusive.

Step 3: Apply the Comparison Test

We compare \(\frac{1}{n \sqrt[n]{n}}\) with \(\frac{1}{n^2}\). For large values of \(n\), the term \(\sqrt[n]{n} = n^{1/n}\) is approximately 1, so \(\frac{1}{n \sqrt[n]{n}} \approx \frac{1}{n^2}\), which is known to converge. Since \(\frac{1}{n \sqrt[n]{n}} < \frac{1}{n^2}\) for all \(n\) large enough, by comparison, the given series converges.

Key Concepts

Root Test

The root test, also known as Cauchy's root test, is a method used to determine the convergence or divergence of infinite series. It's especially useful for series whose general term involves roots. To apply the root test, you calculate the limit of the nth root of the absolute value of the terms of the series. For a series \( \sum a_n \), this means you find \( L = \lim_{n \to \infty} \sqrt[n]{|a_n|} \).
If \( L < 1 \), the series converges absolutely. If \( L > 1 \), the series diverges. If \( L = 1 \), the test is inconclusive, and another test is needed.

In the given problem, the series is \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}} \). By applying the root test, we find that the limit \( L = 1 \), which is inconclusive. Therefore, additional tests, such as the comparison test, are required to reach a conclusion on convergence.

Comparison Test

The comparison test is a powerful tool to determine if a series converges or diverges by comparing it with another series whose behavior is known. The idea is simple:

• If two series \( \sum a_n \) and \( \sum b_n \) are such that \( 0 \leq a_n \leq b_n \) for all \( n \) beyond a certain point, and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• Conversely, if \( \sum a_n \) diverges, then \( \sum b_n \) also diverges given \( a_n \geq b_n \geq 0 \) beyond some point.

In our series \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}} \), we approximate \( \sqrt[n]{n} \approx 1 \) for large \( n \). So \( \frac{1}{n \sqrt[n]{n}} \approx \frac{1}{n^2} \), and since \( \sum \frac{1}{n^2} \) is a convergent p-series, it can be used as a comparison. The inequality \( \frac{1}{n \sqrt[n]{n}} < \frac{1}{n^2} \) for sufficiently large \( n \) shows that our series converges by the comparison test.

P-Series

A p-series is a type of series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a constant. Its convergence depends on the value of \( p \):

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

For example, the series \( \sum \frac{1}{n^2} \) is a p-series with \( p = 2 \), which converges. Recognizing p-series is critical, as they often serve as reference points for comparison tests.

In the case of our original series, we used \( \sum \frac{1}{n^2} \) as the series for comparison. Knowing that it converges (since \( p = 2 \)), helps us determine the convergence of other related series using the comparative approach.