Problem 40 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=2}^{\infty} \frac{n}{(\ln n)^{(n / 2)}} $$

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Identify the Series Type

The given series is \( \sum_{n=2}^{\infty} \frac{n}{(\ln n)^{(n/2)}} \). It is not a p-series or a geometric series directly. We will consider the ratio of terms or growth behavior to determine convergence or divergence.

Step 2: Apply the Ratio Test

Using the Ratio Test, examine the limit of the absolute value of consecutive terms: \[\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} \left| \frac{\frac{n+1}{(\ln(n+1))^{((n+1)/2)}}}{\frac{n}{(\ln n)^{(n/2)}}} \right| \]

Step 3: Simplify the Ratio

Simplifying, we need to calculate:\[\lim_{{n \to \infty}} \frac{n+1}{n} \cdot \left(\frac{\ln n}{\ln(n+1)}\right)^{n/2} \cdot (\ln n)^{-1/2} \cdot (\ln(n+1))^{-1/2}\]Use approximations like \(1+\frac{1}{n} \approx e^{1/n}\) and \(\ln(n+1) \approx \ln n + \frac{1}{n}\) as \(n\) approaches infinity.

Step 4: Evaluate the Limit

Evaluating the limit:- The first factor \( \frac{n+1}{n} \to 1 \).- For the second factor, \( \left( \frac{\ln n}{\ln(n+1)} \right)^{n/2} \approx \left(1 - \frac{1}{n \ln n}\right)^{n/2} \to 1 \) for large \(n\).- The third factor \( (\ln n)^{-1/2} \cdot (\ln(n+1))^{-1/2} \to 0 \) as \( n \to \infty \).Together, this tells us that the limit is 0, which is less than 1.

Step 5: Conclude with the Ratio Test

Since the result of the limit from the Ratio Test is 0, which is less than 1, the series converges by the Ratio Test. Thus, \( \sum_{n=2}^{\infty} \frac{n}{(\ln n)^{(n/2)}} \) converges.

Key Concepts

Ratio Test

The Ratio Test is a handy tool when analyzing infinite series to determine whether they converge or diverge. It uses a simple comparison of the limits of the absolute values of consecutive terms.

To use the Ratio Test, you need to find the limit:\[ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = L \]

• If \( L < 1 \), the series converges.
• If \( L > 1 \) or \( L = \infty \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

In the provided exercise, the Ratio Test revealed that the limit of the ratio of consecutive terms was 0. This result means the series converged, as the limit was less than 1.

Convergence and Divergence

In the context of mathematical series, understanding convergence and divergence is essential. A series converges if the sum of its terms approaches a finite value as the number of terms goes to infinity. On the other hand, a divergent series is one where the sum does not settle into a single value.

The series in the exercise is determined to converge using the Ratio Test. Convergence implies that, though the series has infinitely many terms, they sum up to a specific number. This is a valuable property in mathematical analysis, especially when predicting and estimating the behavior of complex systems. Divergence, conversely, can indicate uncontrolled growth, implying the sum "runs away" to infinity.

Mathematical Series Analysis

When performing mathematical series analysis, it is crucial to start by identifying the type of series you are dealing with. Knowing whether a series is geometric, telescoping, or involves functions like logarithms and exponential terms can guide your choice of methods for determining convergence.

In the exercise, the challenge was to recognize that traditional strategies like checking if it's a simple geometric or p-series wouldn't work. Thus, the Ratio Test became the go-to analysis tool.

• Check for simple types first: geometric series has a constant ratio, and p-series looks like \( \sum \frac{1}{n^p} \).
• If not simple, utilize advanced methods: the Ratio Test, Root Test, or Comparison Test might be appropriate.

Analyzing series is about applying the right test and interpreting results in context, as shown in the solution that decided the convergence nature of the given logarithmic series.