Problem 65 - Essential Calculus

Question

Use the Root Test to determine the convergence or divergence of the series. $$ \sum_{n=2}^{\infty} \frac{(-1)^{n}}{(\ln n)^{n}} $$

Step-by-Step Solution

Answer

The series \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{(\ln n)^{n}}\) converges.

Step 1: Define the Sequence

First, define the sequence for the series as \(a_n= \frac{(-1)^{n}}{(\ln n)^{n}}\). The function of the Root Test is to inspect the behaviour of the \(n\)th root of the absolute value of the \(n\)th term of the series.

Step 2: Apply the Root Test

The Root Test implies that you have to take the limit of the \(n\)th root of absolute value of \(a_n\) as \(n\) approaches infinity. This translates to \( \lim_{n->\infty} |a_n|^{1/n} \). In this case, because \(a_n= \frac{(-1)^{n}}{(\ln n)^{n}}\), \(|a_n| = \frac{1}{|(\ln n)^{n}|}\) and hence, \(|a_n|^{1/n} = \frac{1}{|\ln n|}\).

Step 3: Evaluate the Limit

Evaluate the limit of \(\frac{1}{|\ln n|}\) as \(n\) approaches infinity, which is \( \lim_{n->\infty} \frac{1}{|\ln n|} = 0 \).

Step 4: Conclusion

Since the limit is less than 1, the Root Test states that the series \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{(\ln n)^{n}}\) converges.

Key Concepts

Convergence and Divergence of Series

When it comes to understanding the behavior of infinite series, two fundamental concepts are convergence and divergence. A series is considered convergent if the sum of its terms approaches a specific value as more and more terms are added. Conversely, a series diverges if its terms do not approach a specific limit but instead grow without bound or fluctuate indefinitely.

One of the ways to determine the convergence or divergence of a series is by using the Root Test. This involves examining the nth root of the absolute value of the nth term of the series and determining the limit as n approaches infinity. If this limit is less than 1, the series converges; if it is greater than 1, the series diverges; if it equals 1, the test is inconclusive. In the given exercise, the Root Test helped us conclude that the series is convergent because the calculated limit was zero, which is indeed less than 1.

Limits in Calculus

Limits are a cornerstone of calculus and involve understanding the value that a function or sequence approaches as the input approaches a certain point. They are essential in defining derivatives, integrals, and the behavior of sequences. When applying the Root Test, we are particularly interested in the limit of the nth root of an absolute value of a term in the series as n goes to infinity.

Calculating this limit can often require some tools from calculus, such as L'Hôpital's rule, in case of indeterminate forms, or an understanding of the behavior of functions as variables grow large. In our textbook problem, evaluating the limit of \(\frac{1}{|\ln(n)|}\) as n approaches infinity is straightforward since the natural logarithm of n grows without bound, making the denominator large and the entire fraction approach zero.

Alternating Series

An alternating series is one in which the terms alternate in sign. This means each term is the opposite in sign to the preceding term. In the case of the series in our exercise, \(\frac{(-1)^n}{(\ln n)^n}\), the presence of \( (-1)^n \) means the series will alternate between positive and negative terms.

Understanding alternating series is critical because they often converge under conditions where similar non-alternating series would not. There are specific tests for convergence, like the Alternating Series Test, that apply to these types of series. While we did not utilise this test in our problem since the Root Test was a better fit, it's important to know that different tools and tests exist for handling various series.

Natural Logarithm Properties

The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is approximately 2.71828. It plays a significant role in calculus due to its unique properties, such as the derivative of \(\ln(x)\) being \(\frac{1}{x}\) and the fact that it