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Chapter 11: Problem 28

\(3-32\) Determine whether the series converges or diverges. $$\sum_{n=1}^{\infty} \frac{e^{1 / n}}{n}$$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Express the Series

The series in question is \(\sum_{n=1}^{\infty} \frac{e^{1/n}}{n}\). Here, you will notice the general term is \(a_n = \frac{e^{1/n}}{n}\).
02

Analyze the General Term

Consider the behavior of \(e^{1/n}\) as \(n\to\infty\). As \(n\) becomes very large, \(1/n\) approaches 0, and consequently, \(e^{1/n} \approx 1 + \frac{1}{n}\) by using the first two terms of the Taylor expansion for \(e^x\) at \(x = 0\). Therefore, \(a_n\approx \frac{1 + 1/n}{n} = \frac{1}{n} + \frac{1}{n^2}\).
03

Compare with a Known Series

Observe that \(\frac{1}{n}\) is the general term of the harmonic series, which is known to diverge. The other component \(\frac{1}{n^2}\) of \(a_n\) creates a negligible positive term compared to the divergent part \(\frac{1}{n}\).
04

Apply the Limit Comparison Test

Apply the Limit Comparison Test with the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\). Compute \(\lim_{n \to \infty} \frac{a_n}{b_n}\) where \(a_n = \frac{e^{1/n}}{n}\) and \(b_n = \frac{1}{n}\). This limit is \(\lim_{n \to \infty} \frac{e^{1/n}}{1} = 1\). As the limit is finite and non-zero, it implies that the behavior of \(\sum_{n=1}^{\infty} \frac{e^{1/n}}{n}\) is the same as the harmonic series, and therefore it diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Harmonic Series
When you hear about the harmonic series, think of it as a classic example in the world of series. It's often one of the first infinite series that math students encounter. The harmonic series is written as \( \sum_{n=1}^{\infty} \frac{1}{n} \). The series begins with terms like 1, 1/2, 1/3, and so on.
  • The individual terms get smaller as \( n \) increases.
  • Even though they shrink, the series diverges, meaning the sum grows without bound as \( n \) approaches infinity.
This series is crucial for understanding many concepts in calculus and analysis. Its divergence is surprising to many because the terms appear to get so small.
Limit Comparison Test
The limit comparison test is a handy tool for determining whether an infinite series converges or diverges. It's particularly useful when you have a series that resembles a more familiar one whose behavior you already know.
  • For two series \( \sum a_n \) and \( \sum b_n \), compute \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
  • If this limit is a positive finite constant, both series either converge or diverge together.
In the given exercise, we utilized the harmonic series as \( b_n \), knowing it's divergent. The limit, \( \lim_{n \to \infty} \frac{\frac{e^{1/n}}{n}}{\frac{1}{n}} \), simplifies to 1. Therefore, it confirms the divergence of the given series.
Taylor Expansion
Taylor expansion is a powerful mathematical technique to approximate functions near a specific point. For the exponential function \( e^x \), its Taylor expansion at \( x = 0 \) is particularly useful:
  • \( e^x \approx 1 + x + \frac{x^2}{2!} + \cdots \)
For small values of \( x \), this expansion provides an excellent approximation. In our exercise, using \( x = 1/n \) helped simplify the term \( e^{1/n} \) as it approaches 1 as \( n \) reaches infinity. Thus, the expression \( e^{1/n} \approx 1 + \frac{1}{n} \) was instrumental in breaking down the series term \( \frac{e^{1/n}}{n} \) for further analysis against known series like the harmonic series.

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