Problem 28 - Thomas Calculus

Question

Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{(-\ln n)^{n}}{n^{n}}$$

Step-by-Step Solution

Answer

The series converges by the ratio test.

Step 1: Identify the type of series

The given series is \( \sum_{n=1}^{\infty} \frac{(-\ln n)^{n}}{n^{n}} \). This is a series where each term is given by \( \frac{(-\ln n)^{n}}{n^{n}} \). The presence of \( n^{n} \) in the denominator suggests a comparison to an exponential decay series.

Step 2: Apply the ratio test

The ratio test is applied to determine the convergence of the series. Define \( a_n = \frac{(-\ln n)^{n}}{n^{n}} \). Compute the ratio \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).\[\frac{a_{n+1}}{a_n} = \frac{((-\ln (n+1))^{n+1})/(n+1)^{n+1}}{((-\ln n)^{n})/ n^{n}} = \frac{(-\ln (n+1))^{n+1} \cdot n^n}{(-\ln n)^n \cdot (n+1)^{n+1}}.\]Simplify this expression to find the limit.

Step 3: Simplify the ratio

In simplifying, focus on the dominant terms:\[\left| \frac{(-\ln (n+1))^{n+1} \cdot n^n}{(-\ln n)^n \cdot (n+1)^{n+1}} \right| = \left| \frac{(-\ln (n+1))^{n+1}}{(-\ln n)^n} \right| \cdot \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1}.\]As \( n \to \infty \), consider the limits of each part separately. Notice that: 1. \( \frac{n^n}{(n+1)^n} = \left( \frac{n}{n+1} \right)^n \approx e^{-n/(n+1)} \to e^{-1} \).2. \( \frac{(-\ln (n+1))^{n+1}}{(-\ln n)^n} \) is complicated, but recognize logarithmic growth is slower than polynomial, potentially leading to a diminishing result overall.

Step 4: Conclude with the ratio test outcome

The final outcome of each limit considered in the ratio test tends to zero. Thus, since \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L < 1 \), according to the ratio test, the series converges.

Key Concepts

Ratio Test

The ratio test is a powerful tool in determining whether a series converges or diverges. It involves examining the limit of the absolute value of the ratio of successive terms in a series. For a given series \( \sum a_n \), define the ratio as \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

• If this limit (denoted as \( L \)) is less than 1, the series converges.
• If \( L \) is greater than 1, or if the limit is infinity, the series diverges.
• If \( L \) equals 1, the test is inconclusive.

In the provided exercise, the series under consideration is \( \sum \frac{(-\ln n)^n}{n^n} \). Here, applying the ratio test involves some simplification of the expression for \( \frac{a_{n+1}}{a_n} \). After navigating through the complexities of logarithmic terms and observing the dominant behavior of each term, we find that as \( n \to \infty \), this ratio tends towards zero. Thus, \( L < 1 \), which validates that the series is convergent according to the ratio test.

Exponential Decay

Exponential decay describes a process where quantities decrease at a rate proportional to their current value. Mathematically, this can be represented by a function like \( a_n = \frac{1}{b^n} \), where \( b > 1 \). As \( n \) increases, \( a_n \) trends towards zero very quickly. In the given series, \( \frac{(-\ln n)^n}{n^n} \), the structure of the denominator \( n^n \) mirrors an exponential decay pattern. \( n^n \) grows very rapidly, overpowering any growth in the numerator, and causing the entire fraction to shrink towards zero. This feature of exponential decay can overshadow other terms involved, such as the slow-growing logarithm in the numerator. Individually, exponential terms tend to lead to convergence in series, as their rapid decline tends to make terms small enough relatively fast. When analyzing convergence, recognizing the influence of exponential decay, especially in the denominator, can predict the ultimate behavior of the series in the long run.

Logarithmic Growth

Logarithmic growth is characterized by the slow increase in a function's value as its input becomes larger. The function \( \ln(n) \) exhibits this behavior, growing more gradually than polynomial or exponential functions. In the sequence and series context, \( (-\ln n)^n \) highlights a blend of logarithmic and polynomial attributes. As \( n \) increases, \( \ln(n) \) itself grows slowly, which may seem significant at first glance but does not compete with more dominant terms in a series when \( n \) is large. Even though \( (-\ln n)^n \) appears complex, it is important to recognize its slower rate of increase. When you pair this with a strong exponential decay in the denominator, it ultimately leads to terms that decrease in value. Thus, understanding logarithmic growth in the context of series helps highlight why certain series converge, given how it can only contribute so much to the sequence’s overall behavior.