Question

Use the limit comparison test to determine whether each of the following series converges or diverges. $$ \sum_{n=1}^{\infty}\left(1-e^{-1 / n}\right)\left(\text { Hint: } 1 / e \approx(1-1 / n)^{n}, \text { so } 1-e^{-1 / n} \approx 1 / n .\right) $$

Short answer

The series diverges.

Step-by-step solution

Understanding the series

We are given the series \( \sum_{n=1}^{\infty}\left(1-e^{-1 / n}\right) \). We need to determine if this series converges or diverges using the limit comparison test.

Choose a comparison series

According to the hint, \( 1-e^{-1/n} \approx 1/n \) for large \( n \). Hence, we choose the comparison series \( \sum_{n=1}^{\infty} \frac{1}{n} \), which is the harmonic series. We know that the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges.

Apply Limit Comparison Test

The limit comparison test involves computing the limit: \[\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1-e^{-1/n}}{1/n}.\] Here, \( a_n = 1-e^{-1/n} \) and \( b_n = \frac{1}{n} \). Simplifying our limit expression we have:\[\lim_{n \to \infty} \left( 1-e^{-1/n} \right) \cdot n.\]

Evaluate the limit

To evaluate \( \lim_{n \to \infty} n(1-e^{-1/n}) \), notice that for large \( n \), we can approximate \( 1-e^{-1/n} \approx \frac{1}{n} \). Thus:\[\lim_{n \to \infty} n\left( 1-e^{-1/n} \right) = \lim_{n \to \infty} n\left( \frac{1}{n} \right) = \lim_{n \to \infty} 1 = 1.\]

Interpret the limit comparison result

Since \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \) and this is a positive finite number, the limit comparison test tells us that the behavior of the series \( \sum a_n \) is the same as \( \sum b_n \). Since the harmonic series \( \sum \frac{1}{n} \) diverges, the given series \( \sum_{n=1}^{\infty}\left(1-e^{-1 / n}\right) \) also diverges.

Key concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool in calculus for determining the convergence or divergence of an infinite series. It works by comparing an unknown series with a more familiar series whose behavior is already understood. This test is particularly useful when the series terms resemble each other as they approach infinity.

Here’s how the limit comparison test works:

• Suppose you have two series, \( \sum a_n \) and \( \sum b_n \).
• To apply the test, compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
• If this limit is a positive finite number, both series converge or both diverge.

Understanding the outcome relies on knowing the behavior of the comparison series. In our exercise, we use the harmonic series as our comparison series. Since the harmonic series is known to diverge, if our test gives a finite positive limit, our original series must also diverge.

Harmonic Series

The harmonic series is one of the most famous and classic examples in calculus known for its divergence. It is expressed as \( \sum_{n=1}^{\infty} \frac{1}{n} \). Despite the terms getting smaller and smaller as \( n \) increases, they do not decrease fast enough for the series to converge.

Here's why it's important:

• The divergence of the harmonic series forms a common foundational point for analyzing other series.
• It serves as a benchmark comparison for applying the limit comparison test.
• Understanding the behavior of the harmonic series helps in quickly identifying the divergence of similar series.

In our exercise, recognizing that the harmonic series diverges allowed us to conclude the divergence of the series \( \sum_{n=1}^{\infty} (1 - e^{-1/n}) \) when using the limit comparison test.

Series Divergence

Divergence in the context of series means that as you sum more and more of the series' terms, the result keeps increasing without bound. Essentially, the total sum does not add up to any finite number. Identifying divergence involves applying tests and comparing series efficiently.

Key points to understand series divergence include:

• Recognizing that if a series is divergent, its partial sums grow indefinitely.
• Tests like the limit comparison test help us establish divergence by comparing with already known divergent series.
• Simplifications and approximations play a crucial role when evaluating limits to check for divergence.

In our example, finding that the limit \( \lim_{n \to \infty} \frac{1-e^{-1/n}}{1/n} = 1 \) indicates that our series behaves similarly to the divergent harmonic series, confirming its divergence.