91Ó°ÊÓ

When analyzing series, getting a handle on the behavior of the general term is often the first step. For the series \( \sum_{n=1}^{\infty} \frac{e^{1/n}}{n} \), we begin by examining the term \( e^{1/n} \). As \( n \) becomes larger, \( \frac{1}{n} \) gets closer to zero. This leads us to use the series expansion for the exponential function:

• \( e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

For a small \( x = \frac{1}{n} \), the terms after \( x \) become very tiny, so it's reasonable to approximate \( e^{1/n} \approx 1 + \frac{1}{n} \). This approximation simplifies our series term to \( \frac{1}{n} + \frac{1}{n^2} \).
This simplification is useful because it allows us to analyze the series by separating it into two more familiar series: \( \sum \frac{1}{n} \) and \( \sum \frac{1}{n^2} \).
This process of finding an approximation gives us a clearer picture of what the series resembles, helping us decide whether it converges or diverges.

The harmonic series, represented as \( \sum_{n=1}^{\infty} \frac{1}{n} \), is a series that is well-known in mathematics. It may seem simple because each term gets smaller, yet the sum keeps increasing without bound.
The behavior of the harmonic series provides a crucial insight: it diverges. This means that if a series behaves like the harmonic series as \( n \to \infty \), there's a good probability the series also diverges.
In our approximation of \( \sum \frac{e^{1/n}}{n} \), we find a segment that strongly echoes the harmonic series, \( \sum \frac{1}{n} \).
Whenever we approach convergence questions, examining the resemblance of a series to the harmonic series can be a decisive factor for determining divergence. This is essential in making a comparison and drawing conclusions about the original series's behavior.

Throughout mathematics, a p-series is defined as a series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Depending on the value of \( p \), the series may converge or diverge:

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

In studying our problem's approximation \( \frac{1}{n} + \frac{1}{n^2} \), the term \( \frac{1}{n^2} \) aligns with a p-series where \( p = 2 \).
This segment of the series converges because \( 2 > 1 \).
Understanding p-series helps us aggregate the convergence behavior of separate terms in a more complex series, aiding our overall analysis.
By knowing how the series behaves with respect to p-series, we lay the groundwork for comparing its segments thoroughly.

The comparison test is a reliable method in determining the convergence of series by juxtaposing it with another series whose behavior (converging or diverging) is already known.
If the terms of a series are greater than or equal to a divergent series and the original series behaves similarly, it will diverge.
Alternatively, if the terms are less than those of a convergent series, it too will converge.
In the case of \( \sum \frac{e^{1/n}}{n} \), we found it could be approximated as a mix of two known series: the divergent harmonic series \( \sum \frac{1}{n} \) and the convergent p-series with \( p = 2 \), \( \sum \frac{1}{n^2} \).
By focusing on the comparison to the harmonic series—since \( \frac{e^{1/n}}{n} \approx \frac{1}{n} \)—we apply the comparison test to discern the overall behavior. Despite having a convergent p-series component, the series diverges because the dominant term relates to the divergent harmonic series.
This reasoning underscores the importance of isolating dominant behaviors in a composite series when using the comparison test.