Problem 26 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{n} $$

Step-by-Step Solution

Answer

The series diverges because the terms do not approach zero.

Step 1: Recognize the Type of the Series

The series given is \( \sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n} \). This series can be viewed as the sum of terms \( a_n = \left(1+\frac{1}{n}\right)^{n} \). These terms are not a straightforward geometric or p-series, nor can they immediately be classified as a telescoping series or an alternating series. We need to analyze the behavior of the terms as \( n \) becomes very large.

Step 2: Use Limit Comparison with Known Functions

Compute the limit of the expression inside the summation: \( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \). Using the known limit \( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \), where \( e \) is the base of the natural logarithm, we conclude that \( a_n \approaches e \) as \( n \) approaches infinity.

Step 3: Apply Divergence Test

The Divergence Test states that if \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum_{n=1}^{\infty} a_n \) diverges. Since \( \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = e \), which is a constant \( eq 0 \), we conclude that the series \( \sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n} \) diverges.

Key Concepts

Divergence Test

The Divergence Test, also known as the Test for Divergence, is a fundamental tool used to determine if a series diverges. The essence of this test is quite simple. It helps us figure out if the individual terms in a series tend to zero as the number increases, which is a necessary condition for the series to have a chance of converging.

Here's how it works:

• If the limit of the terms of a series, as they approach infinity, is not zero or does not exist, the series must diverge.
• Mathematically, for a series \(\sum_{n=1}^{\infty} a_n\), if \(\lim_{n \to \infty} a_n eq 0\), then the series diverges.
• It is important to remember that if the terms do go to zero, the series might still diverge. The test only confirms divergence, but not convergence.

For the given series \(\sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n}\), we find that as \(n\) approaches infinity, the limit of the terms is \(e\), the base of natural logarithms, which is not zero. Therefore, the series diverges.

Limit Comparison

The Limit Comparison Test is used to compare a difficult-to-evaluate series with a simpler one, whose behavior is known. Essentially, this test helps ascertain whether two series behave similarly by comparing the limits of their terms.

Here's how you apply the Limit Comparison Test:

• You compare the terms of your series, \(a_n\), to the terms of another series, \(b_n\), whose convergence is already known.
• Compute \(L = \lim_{n \to \infty} \frac{a_n}{b_n}\).
• If \(L\) is a finite number greater than zero, both series either converge or diverge together.
• If \(L\) equals zero or infinity, the test gives no information.

For the series \(\left(1+\frac{1}{n}\right)^{n}\), the limit comparison was more implicit. We used it in conjunction with known limits of elementary functions to decide that the terms tend towards \(e\), thus leading us to apply the Divergence Test.

Natural Logarithm

Natural logarithms are logarithms with base \(e\), where \(e \approx 2.718\). This is a fundamental constant in mathematics and appears frequently in calculus and analysis.

Why are natural logarithms important?

• They are the inverses of exponential functions with base \(e\).
• They simplify expressions involving exponential growth, like \(\left(1+\frac{1}{n}\right)^{n}\).
• They arise naturally in problems of growth and decay, such as population dynamics or radioactivity.

In the context of the provided series, \(e\) itself is a crucial part as it forms the predicted limit of the individual series terms. Understanding this constant helps us evaluate the convergence of series where terms involve exponential behaviors. This is why identifying limits like \(\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = e\) is vital for using the divergence and comparison tests effectively.