Problem 25 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty}\left(\frac{n}{3 n+1}\right)^{n} $$

Step-by-Step Solution

Answer

The series converges by the root test.

Step 1: Analyze the General Term

Consider the general term of the series: \( a_n = \left(\frac{n}{3n+1}\right)^n \). As \( n \to \infty \), evaluate how \( \frac{n}{3n+1} \) behaves. This can be rewritten as \( \frac{n}{3n+1} = \frac{1}{3 + \frac{1}{n}} \). As \( n \to \infty \), \( \frac{1}{n} \to 0 \), so \( \frac{1}{3 + \frac{1}{n}} \to \frac{1}{3} \). Therefore, \( a_n \) asymptotically behaves like \( \left(\frac{1}{3}\right)^n \).

Step 2: Apply the Root Test

The root test, which is useful for series \( \sum a_n \), involves finding \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \). Apply this to \( a_n = \left(\frac{n}{3n+1}\right)^n \):\[ \lim_{n \to \infty} \sqrt[n]{\left(\frac{n}{3n+1}\right)^n} = \lim_{n \to \infty} \frac{n}{3n+1} = \frac{1}{3} \]Since the limit \( \frac{1}{3} < 1 \), by the root test, the series converges.

Step 3: Conclusion

Based on the root test result, since \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = \frac{1}{3} < 1 \), the series \( \sum_{n=1}^{\infty}\left(\frac{n}{3 n+1}\right)^{n} \) is absolutely convergent by the root test.

Key Concepts

Root Test

The root test is a valuable tool for assessing the convergence of an infinite series. Specifically, it examines the nth root of the absolute value of the general term of a series. Here's how it works:

• Consider a series with general term \( a_n \).
• Find the limit \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \).
• If this limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

In our exercise, applying the root test to the series \( \sum_{n=1}^{\infty}\left(\frac{n}{3n+1}\right)^{n} \), we calculated:\[ \lim_{n \to \infty} \sqrt[n]{\left(\frac{n}{3n+1}\right)^n} = \lim_{n \to \infty} \frac{n}{3n+1} = \frac{1}{3} \]Since \( \frac{1}{3} < 1 \), the root test confirms that this series is convergent.

Infinite Series

An infinite series is simply the sum of infinitely many terms. These terms are derived from a sequence of numbers. The series can take different forms, such as geometric or arithmetic series, and understanding their convergence is essential.
For a series to converge means that, as you add more and more terms, the sum approaches a specific finite value. Conversely, if a series diverges, the sum does not settle at a finite limit.
When working with infinite series, several tests are available to examine whether the series converges or diverges. These include the root test, ratio test, and comparison test among others. Choosing the right one depends on the series structure.
In our case, the series was tested for convergence using the root test, which provided clarity and led to a decisive conclusion.

Asymptotic Behavior

Asymptotic behavior is a concept used to describe how a function behaves as the input approaches a certain point, often infinity. In the context of series, it helps in visualizing the behavior of the sequence terms as they get very large, which is crucial in studying convergence.
To explore this, consider the function \( f(n) = \frac{n}{3n+1} \). As \( n \to \infty \), the term \( \frac{1}{n} \to 0 \), simplifying the expression to behave like \( f(n) \approx \frac{1}{3} \). Such approximation indicates how the terms \( a_n = \left(\frac{n}{3n+1}\right)^n \) in the series approach a simple predictable pattern, which in turn informs convergence testing.
Understanding asymptotic behavior allows us to anticipate the results of convergence tests, such as the root test employed in this exercise. By recognizing that \( a_n \) behaves like \( \left(\frac{1}{3}\right)^n \), it became apparent that the series converged as \( \frac{1}{3} < 1 \). This foresight into the series' asymptotic nature facilitated an efficient and clear application of the root test.