Problem 15 - Calculus Single Variable

Question

State what conclusion, if any, may be drawn from the Divergence Test. $$ \sum_{n=1}^{\infty} n^{1 / n} $$

Step-by-Step Solution

Answer

The series diverges by the Divergence Test since the limit of the terms is not zero.

Step 1: Identify the sequence terms

To apply the Divergence Test, we first identify the sequence terms of the series. In this exercise, the terms are given by \( n^{1/n} \).

Step 2: Apply the Divergence Test

The Divergence Test states that if the limit of the sequence terms \( a_n \) (in this case \( n^{1/n} \)) is not equal to zero, then the series \( \sum_{n=1}^{\infty} a_n \) diverges.

Step 3: Compute the limit of the sequence

We compute \( \lim_{n \to \infty} n^{1/n} \). Let's simplify this: \( n^{1/n} = e^{\ln(n^{1/n})} = e^{(1/n) \cdot \ln n} \). This is equivalent to finding \( \lim_{n \to \infty} e^{(1/n) \cdot \ln n} \).

Step 4: Evaluate the exponent limit

Calculate \( \lim_{n \to \infty} \frac{\ln n}{n} \). Using L'Hôpital's Rule as both the numerator and denominator go to infinity, we differentiate the numerator and the denominator: \( \frac{d}{dn}(\ln n) = \frac{1}{n} \) and \( \frac{d}{dn}(n) = 1 \). Then the limit is \( \lim_{n \to \infty} \frac{1/n}{1} = \lim_{n \to \infty} \frac{1}{n} = 0 \).

Step 5: Find the limit of the original expression

Since \( \lim_{n \to \infty} \frac{\ln n}{n} = 0 \), we get \( \lim_{n \to \infty} e^{(1/n) \cdot \ln n} = e^0 = 1 \).

Step 6: Conclude using the Divergence Test

Since \( \lim_{n \to \infty} n^{1/n} = 1 eq 0 \), by the Divergence Test, we conclude that the series \( \sum_{n=1}^{\infty} n^{1/n} \) diverges.

Key Concepts

Limit of a Sequence

When approaching problems involving series, we often need to understand the concept of the limit of a sequence. The limit of a sequence refers to the value that the terms of a sequence approach as the index, usually denoted by \( n \), becomes very large.

In the exercise, the sequence in question is \( n^{1/n} \). To find if this sequence converges to a limit, we calculate \( \lim_{n \to \infty} n^{1/n} \). Simplifying, we have:

• Express using an exponent: \( n^{1/n} = e^{(1/n) \cdot \ln n} \)
• Target: \( \lim_{n \to \infty} e^{(1/n) \cdot \ln n} \)

Understanding these sequence limits is crucial, as they help determine whether such series might converge or diverge when summed across infinity. The specific calculation involves examining how the terms of the sequence behave as \( n \) increases towards infinity, guiding us in our analysis and conclusion for the given series.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in evaluating limits of indeterminate forms, such as \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \). This rule states that if these forms exist, the limit of functions can be found by differentiating the numerator and denominator separately.

In our calculation for \( \lim_{n \to \infty} \frac{\ln n}{n} \), we encounter an indeterminate form \( \frac{\infty}{\infty} \). This is where L'Hôpital's Rule can be applied:

• Differentiate the numerator: \( \frac{d}{dn}(\ln n) = \frac{1}{n} \)
• Differentiate the denominator: \( \frac{d}{dn}(n) = 1 \)
• Evaluate the new limit: \( \lim_{n \to \infty} \frac{1/n}{1} = \lim_{n \to \infty} \frac{1}{n} = 0 \)

This tells us that as \( n \) tends towards infinity, the \( \frac{\ln n}{n} \) approaches zero. With practice, using L'Hôpital's Rule becomes more intuitive, simplifying the evaluation of complex limits.

Series Divergence

The concept of series divergence plays a crucial role when working with infinite series. A series diverges if the sum of its sequence terms does not settle towards a finite limit.

The Divergence Test helps identify such divergence by checking the behavior of the sequence terms' limits:

• If \( \lim_{n \to \infty} a_n eq 0 \), the series \( \sum_{n=1}^{\infty} a_n \) diverges.
• In our exercise, the limit of the sequence \( \lim_{n \to \infty} n^{1/n} = 1 \).
• Since this is not zero (\( 1 eq 0 \)), by the Divergence Test, \( \sum_{n=1}^{\infty} n^{1/n} \) diverges.

Recognizing divergence is key in series analysis, as it allows us to conclude behaviors of sums that extend to infinity, informing the feasibility and convergence of real-world applications modeled by such series.