Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{n=2}^{\infty}\left(1-\frac{1}{n}\right)^{n}\)

Step-by-Step Solution

Answer

The series diverges.

Step 1: Identifying the Type of Series

Recognize that the series is \(\sum_{n=2}^{\infty}\left(1-\frac{1}{n}\right)^{n}\) and try to determine its convergence or divergence. Evaluate the function \(a_n = \left(1-\frac{1}{n}\right)^{n}\) as \(n\) approaches infinity.

Step 2: Expressing the Series in Terms of Exponential

Recall the exponential limit \(\left(1-\frac{1}{n}\right)^{n} \approx e^{-1}\) as \(n\rightarrow\infty\) using the approximation \(\left(1+\frac{x}{n}\right)^{n} \approx e^{x}\) for \(x=-1\). Hence, \(a_n\) converges to \(e^{-1}\).

Step 3: Testing Convergence with Limit Comparison Test

Use the limit comparison test with a known convergent or divergent series. Consider the series \(b_n = \frac{1}{n}\). Compute the limit \(\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} n\left(1-\frac{1}{n}\right)^n.\) Simplify this expression to see if it leads to a constant.

Step 4: Calculating the Limit

To compute \(\lim_{n\to\infty} n\left(1-\frac{1}{n}\right)^n\), start by using the property \(\left(1-\frac{1}{n}\right)^n = e^{-1} e^{-a_n}\) where the correction term \(e^{-a_n} \approx 1-\frac{1}{n} \). Therefore, \(n\left(1-\frac{1}{n}\right)^n \approx n\cdot\frac{1}{ne}\) converges to \(\frac{1}{e}\).

Step 5: Concluding the Convergence

Since \(\lim_{n\rightarrow\infty} \frac{a_n}{b_n} = \frac{1}{e}\) which is a non-zero constant, \(\sum_{n=2}^{\infty} a_n\) and \(\sum_{n=2}^{\infty} \frac{1}{n}\) either both converge or both diverge. Given that \(\sum_{n=2}^{\infty} \frac{1}{n}\) diverges, the original series \(\sum_{n=2}^{\infty} \left(1-\frac{1}{n}\right)^{n}\) also diverges.

Key Concepts

Limit Comparison Test

The limit comparison test is a powerful tool to determine if an infinite series converges or diverges.

When faced with a series \(\sum a_n\), comparing it to another series \(\sum b_n\) that is easier to analyze can simplify the problem. Here's how it works:

• Choose a series \(\sum b_n\) that is similar in form to \(\sum a_n\).
• Ensure you know whether \(\sum b_n\) is convergent or divergent.
• Compute the limit \(L = \lim_{{n \to \infty}} \frac{a_n}{b_n}\).
• If \(L\) is a positive and finite number, and \(L eq 0\), then both \(\sum a_n\) and \(\sum b_n\) will either both converge or both diverge.

In this exercise, we used the series \(b_n = \frac{1}{n}\). We found that the original series and this simple \(\sum \frac{1}{n}\), which is known to diverge (the harmonic series), led to the conclusion that the series \(\sum \left(1-\frac{1}{n}\right)^{n}\) also diverges.

Exponential Limit

An exponential limit is a particular calculation in which an expression approaches an exponential function's base value as a variable approaches infinity.

In the given series, we recognize that the general term \(a_n = \left(1-\frac{1}{n}\right)^{n}\) is linked to an exponential limit. The key result used here is:

• When \(n\) approaches infinity, \(\left(1+\frac{x}{n}\right)^{n}\) approaches \(e^x\).
• Setting \(x = -1\), we find \(\left(1-\frac{1}{n}\right)^n \approx e^{-1}\).

This simplification shows that each term \(a_n\) approximates \(e^{-1}\) for large values of \(n\). This exponential limit helps us simplify complex expressions and is pivotal in our test for convergence or divergence. By understanding these exponential behaviors, students can appreciate how certain forms transform into familiar constants like \(e\).

Divergent Series

A divergent series is one in which the sum of its infinite terms does not lead to a finite number.

Unlike convergent series, these continue to increase without bound or oscillate indefinitely.

In this exercise, after performing the limit comparison test, we concluded that the series \(\sum \left(1-\frac{1}{n}\right)^{n}\) diverges. Here's why:

• We compared it with the harmonic series \(\sum \frac{1}{n}\), a classic example of a divergent series.
• The comparison provided a non-zero and finite constant result in the limit comparison test.
• Since the harmonic series diverges, and our original series has the same divergence characteristics, it must also diverge.

Understanding divergence is crucial, as it highlights series that do not settle to a fixed value. This concept is fundamental in mathematical analysis, helping distinguish which problems require additional techniques or considerations when summing infinite series.