Problem 42 - 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{\ln n}{\sqrt{n} e^{n}}\end{equation}

Step-by-Step Solution

Answer

The series converges by the Limit Comparison Test with the geometric series \( \sum \frac{1}{e^n} \).

Step 1: Determine the Terms Type

The given series is \( \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} \). This series is a sum of terms that are decreasing and positive for \( n \geq 1 \). We will check if these terms approach zero to consider the convergence.

Step 2: Apply the Limit Comparison Test

To determine convergence, we can use the Limit Comparison Test. Choose a comparison series \( b_n = \frac{1}{e^n} \) which is a geometric series with a common ratio \( r = \frac{1}{e} < 1 \) and is known to converge.

Step 3: Compute the Limit for Comparison

Calculate \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\ln n}{\sqrt{n}} = \lim_{n \to \infty} \frac{\ln n}{\sqrt{n}} \). We can analyze this limit by noting that \( \ln n \) grows slower than \( \sqrt{n} \) so this limit approaches \( 0 \) as \( n \to \infty \).

Step 4: Conclude using Limit Comparison Test

Since \( \frac{\ln n}{\sqrt{n}} \) approaches zero, this implies \( \lim_{n \to \infty} \frac{a_n}{b_n} = 0 \). Hence, by the Limit Comparison Test, since \( \sum b_n = \sum \frac{1}{e^n} \) converges, \( \sum a_n = \sum \frac{\ln n}{\sqrt{n} e^{n}} \) also converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a useful method to determine whether a series converges or diverges. It's particularly handy when direct comparison with another series is suitable but needs further simplification. To use this test, you compare the series in question, say \(a_n\), with another series \(b_n\) that you already know the convergence behavior of. Here's how it works in simple steps:- You select a series \(b_n\) such that it's easy to determine whether it converges or diverges.- Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).- If this limit is a positive finite number, then both series either converge or diverge together.In our case, the given series is \(a_n = \frac{\ln n}{\sqrt{n} e^{n}}\). We compare it with the series \(b_n = \frac{1}{e^n}\). This comparison works because \(b_n\) is a geometric series that converges, given its common ratio \(r = \frac{1}{e} < 1\). By evaluating \( \lim_{n \to \infty} \frac{a_n}{b_n} \), we found that the limit approaches zero. Hence, by the Limit Comparison Test, \( \sum a_n \) also converges.This method is particularly useful because it reduces a potentially complex convergence analysis to a simpler comparison.

Geometric Series

A geometric series is one of the most straightforward types of series. Its terms form a sequence where each term is a constant multiple of the previous one. A geometric series takes the form \(a + ar + ar^2 + ar^3 + \ldots\), or more generally as \(\sum_{n=0}^{\infty} ar^n\), where \(a\) is the first term and \(r\) is the common ratio.

Convergence of Geometric Series

A key point about geometric series is that they have a simple criterion for convergence:- If \(|r| < 1\), the series converges.- If \(|r| \geq 1\), the series diverges.The geometric series is crucial in the context of the original problem because our comparison series \(b_n = \frac{1}{e^n}\) has \(r = \frac{1}{e}\). Since \(\frac{1}{e} < 1\), it converges. These properties make geometric series an excellent choice for use in convergence tests like the Limit Comparison Test.

Convergence and Divergence

In the study of series, understanding whether a series converges or diverges is essential. This primarily tells us if the series adds up to a finite value (converges) or not (diverges).

• Convergent Series: A series \(\sum a_n\) converges if the sequence of its partial sums tends to a finite limit as you keep adding terms. This means the series settles at a specific value.
• Divergent Series: Conversely, a series diverges if its partial sums do not approach any finite limit. It may grow indefinitely, oscillate, or produce non-converging results.

In our exercise, by using the Limit Comparison Test with a geometric series, we proved that \( \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} \) is convergent. We analyzed the behavior of the terms and successfully demonstrated that they sum to a finite limit. This understanding is pivotal in calculus as it helps differentiate between series that are calculable and those that are not.