Problem 35 - Calculus: Early Transcendentals

Question

Use any test to determine whether the series is absolutely convergent, conditionally convergent, or divergent. \( \displaystyle \sum_{n = 2}^{\infty} \left( \frac {n}{\ln n} \right) ^n \)

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Recognize the Type of Series

The series given is \( \sum_{n = 2}^{\infty} \left( \frac{n}{\ln n} \right)^n \). This appears to be an exponential series, which makes it suitable for comparison tests.

Step 2: Consider the Comparison Test

To apply the Comparison Test, we need to compare \( \left( \frac{n}{\ln n} \right)^n \) to a known divergent series. We can compare it to \( n^n \). Notice that \( \frac{n}{\ln n} > 1 \) for \( n \geq 2 \). Therefore, \( \left( \frac{n}{\ln n} \right)^n > n^n \). Since \( \sum_{n=2}^{\infty} n^n \) is a divergent series, by the Comparison Test, \( \sum_{n=2}^{\infty} \left( \frac{n}{\ln n} \right)^n \) also diverges.

Key Concepts

Comparison Test

The Comparison Test is a valuable tool in determining whether a series converges or diverges. It's a relatively straightforward method when you can identify another series that you already know the behavior of—meaning you understand whether it converges or diverges. Here's how it works in simple terms:

• If you have a series \(\sum a_n\) and you want to know if it converges, you find a series \(\sum b_n\) that you already know converges, and you show that \(0 \leq a_n \leq b_n\) for all n.


• Conversely, if you want to show divergence, you find a known divergent series \(\sum b_n\) and prove \(a_n \geq b_n > 0\) for all n.

This was the method applied in the original solution for our exercise, where the term \(\left( \frac{n}{\ln n} \right)^n\) was compared to the known divergent series \(n^n\). Since \(\left( \frac{n}{\ln n} \right)^n\) is greater than \(n^n\) when n is equal to or greater than 2, and because \(n^n\) is divergent, the use of the Comparison Test here effectively demonstrates that the series in question is also divergent.

Divergent Series

Understanding what makes a series divergent is crucial for various mathematical contexts. A series is called divergent if, as you continue to add its terms together indefinitely, the sum increases without bound, or does not approach a specific finite limit.

• In simpler terms, no matter how many of the terms you add up, you won't get a stable number; the sum either grows larger and larger or fluctuates.


• Some common examples of divergent series include the harmonic series \(\sum \frac{1}{n}\) and any series where the individual terms do not get smaller fast enough to approach zero.

The series in this exercise, represented by \(\sum_{n=2}^{\infty} \left( \frac{n}{\ln n} \right)^n\), was shown to diverge using the Comparison Test, which identified that the terms of the series outpaced those of the already-known divergent series \(n^n\). This means as you sum up the terms from 2 to infinity, the total sum tends towards infinity rather than converging to a finite number.

Exponential Series

An exponential series is characterized by terms where the variable is in the exponent. In a more general sense, any series of the form \(a^n\) is exponential if a is a constant and n is a variable that changes over the terms of the series. Here are some insights into handling exponential series:

• These series often grow very quickly, especially if the base is greater than 1, making them candidates for divergence.


• The exponential nature means even small increases in n can lead to significant increases in term size, which, in turn, affects convergence.

For our original exercise, the expression \(\left( \frac{n}{\ln n} \right)^n\) is an example of an exponential series since the base \(\frac{n}{\ln n}\) is raised to the power of n. This rapid increase in term size contributes to its divergence, aligning with the knowledge that exponential series often do not converge if the base exceeds 1 as n increases.