Problem 31 - Thomas Calculus

Question

Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{e^{n}}{n^{e}}$$

Step-by-Step Solution

Answer

The series diverges by the Ratio Test.

Step 1: Understand the Series

The given series is \( \sum_{n=1}^{\infty} \frac{e^{n}}{n^{e}} \). We need to determine if this series converges or diverges.

Step 2: Choose a Convergence Test

We will use the Ratio Test to determine convergence. This test is suitable for series where the terms involve powers or factorials, like \( e^n \) in our series.

Step 3: Apply the Ratio Test

For the series \( \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{e^{n}}{n^{e}} \), compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Here, \( a_n = \frac{e^{n}}{n^{e}} \) and \( a_{n+1} = \frac{e^{n+1}}{(n+1)^{e}} \).

Step 4: Simplify the Ratio

Compute \( \frac{a_{n+1}}{a_n} = \frac{e^{n+1} \cdot n^e}{e^n \cdot (n+1)^e} = e \cdot \left(\frac{n}{n+1}\right)^e \).

Step 5: Evaluate the Limit

Evaluate the limit \( \lim_{n \to \infty} e \cdot \left(\frac{n}{n+1}\right)^e \). As \( n \to \infty \), \( \left(\frac{n}{n+1}\right)^e \to 1 \), hence \( \lim_{n \to \infty} e \cdot 1 = e > 1 \).

Step 6: Conclusion from the Ratio Test

Since the limit \( e > 1 \), the Ratio Test indicates that the series diverges.

Key Concepts

Ratio Test

The Ratio Test is a valuable tool when determining whether a series converges or diverges. It's especially useful for series that contain powers or factorial expressions. The basic idea involves examining the ratio between successive terms in the series.

To apply the Ratio Test, you calculate the limit:

• Let \(a_n\) be the general term of your series.
• Compute \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

The results of this limit can help decide:

• If the limit is less than 1, the series converges.
• If the limit equals 1, the test is inconclusive.
• If the limit is greater than 1, the series diverges.

In our original exercise, the series diverges because the limit is greater than 1, specifically it equals \(e\), which indicates divergence according to the test.

Power Series

Power series are a type of infinite series where each term contains a form of a variable raised to a power. They are a significant concept in calculus and analysis, often appearing in functions expressed as an infinite sum.

A general power series looks like this:

• \(\sum_{n=0}^{\infty} c_n (x-a)^n\)

Here:

• \(c_n\) are coefficients or constants.
• \(x\) is the variable in question.
• \(a\) is the center of the series.

Power series converge within a particular radius around \(a\), known as the radius of convergence. Understanding this can be vital for analyzing functions and approximating them.In the context of the original exercise, although it's not about a power series, recognizing series forms helps in choosing the right convergence tests like the Ratio Test to understand their behavior.

Limit Evaluation

Evaluating limits is a crucial step in applying convergence tests like the Ratio Test. Essentially, it helps to determine the behavior of sequences or functions as the input approaches a certain value, often infinity.

When evaluating limits for series, it's essential to remember:

• Understanding growth rates of polynomial, exponential, and logarithmic functions can provide quick insights.
• Common techniques include L'Hôpital's rule, algebraic manipulation, or known limit laws.

In the original problem, we evaluated \( \lim_{n \to \infty} e \cdot \left(\frac{n}{n+1}\right)^e \). The key insight was recognizing \( \left(\frac{n}{n+1}\right)^e \rightarrow 1 \) as \(n\) becomes very large, because the dominant terms in the numerator and denominator cancel each other out. Approaching limits methodically and patiently can clear up the convergence question of complex series.