Problem 25 - Calculus: Early Transcendentals

Question

Determine whether the series converges or diverges. \( \displaystyle \sum_{n = 1}^{\infty} \frac {e^n + 1}{ne^n + 1} \)

Step-by-Step Solution

Answer

The series diverges.

Step 1: Understand the Series

We are given the series \( \sum_{n = 1}^{\infty} \frac {e^n + 1}{ne^n + 1} \). Our goal is to determine whether this series converges or diverges. To do this, we will analyze the behavior of the terms as \( n \to \infty \).

Step 2: Check the Terms

Examine the expression for the general term in the series: \( a_n = \frac {e^n + 1}{ne^n + 1} \). Notice that both the numerator and denominator are dominated by their exponential terms as \( n \) grows, making them approximate \( e^n \) and \( ne^n \), respectively.

Step 3: Simplify the Expression

Simplify \( a_n = \frac {e^n}{ne^n} \) as \( n \to \infty \), which simplifies to \( \frac {1}{n} \) after canceling \( e^n \) from the numerator and denominator. This shows that for large \( n \), the terms of the series behave like \( \frac{1}{n} \).

Step 4: Apply the Divergence Test

The series \( \sum \frac{1}{n} \) (the harmonic series) is a well-known divergent series. If \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n} = 0 \), we cannot conclude convergence yet. However, since the behavior is identical to a divergent series, the limit comparison test may be helpful.

Step 5: Use the Limit Comparison Test

Compare \( a_n = \frac {e^n + 1}{ne^n + 1} \) with \( b_n = \frac{1}{n} \). Compute \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{e^n + 1}{ne^n + 1} \cdot n = \lim_{n \to \infty} \frac{e^n + 1}{e^n + \frac{1}{n}} = 1 \). Since this limit is positive and finite, both series either converge or diverge together.

Step 6: Conclusion

The series \( \sum \frac{1}{n} \) is known to diverge. Hence, by the limit comparison test, the given series \( \sum_{n = 1}^{\infty} \frac {e^n + 1}{ne^n + 1} \) also diverges.

Key Concepts

Divergence Test

The Divergence Test, also known as the nth-term test for divergence, is a simple yet powerful tool in determining whether a series converges or diverges. It's the first check to perform when exploring the behavior of an infinite series.

Here's how it works:

• If the limit of the terms of a series, as n approaches infinity, does not equal zero, then the series must diverge.
• Conversely, if the limit equals zero, the test is inconclusive for convergence, and we must apply other tests to reach a conclusion.

For the series in question: \[ a_n = \frac{e^n + 1}{ne^n + 1} \]As explained in the solution steps, simplifying this term for large n shows that it resembles \(\frac{1}{n}\) in behavior.

Though the divergence test itself doesn't prove convergence here since \(\lim_{n \to \infty} a_n = 0\), it sets the stage for further analysis with more sophisticated methods. This is why it's important to transition to the limit comparison test.

Limit Comparison Test

The Limit Comparison Test is invaluable when the Divergence Test does not provide a definitive answer. It allows us to compare a complicated series with a simpler, well-known series.

• To use this test, we find a series \(b_n\) such that \(b_n\) approximates our series' terms \(a_n\) for large n.
• We compute the limit: \[\lim_{n \to \infty} \frac{a_n}{b_n}\]and if this limit is a positive, finite number, both series either converge or diverge together.

In our example, we compared \[a_n = \frac{e^n + 1}{ne^n + 1}\]with the simpler \[b_n = \frac{1}{n}\]The limit calculation \[\lim_{n \to \infty} \frac{a_n}{b_n} = 1\]indicates that both follow the same convergence behavior. Given that \(\sum \frac{1}{n}\) is a divergent series, our original series will diverge too.

Harmonic Series

The Harmonic Series is a classic example in mathematics, often serving as a benchmark in series convergence discussions. It's represented by:\[\sum_{n=1}^{\infty} \frac{1}{n}\]This series is perhaps most famous for its surprising divergence, despite the terms approaching zero as n increases.

• The harmonic series diverges because the terms decline at a rate too slow for the series to sum to a finite number.
• When working with series like \[\sum_{n=1}^{\infty} \frac{e^n + 1}{ne^n + 1}\],identifying its resemblance to the harmonic series can be instrumental.

By using the harmonic series as a comparative tool, as we did in the limit comparison test, we can conveniently demonstrate that certain complex series diverge. Understanding why the harmonic series diverges can provide deep insights into the behavior of seemingly related series.