Problem 25 - Calculus Single Variable

Question

In each of Exercises \(25-34,\) use the Root Test to determine the convergence or divergence of the given series. \(\sum_{n=1}^{\infty} n^{-n / 2}\)

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} n^{-n/2} \) converges by the Root Test.

Step 1: Understand the Root Test

The Root Test states that for the series \( \sum a_n \), we compute the limit \( L = \lim_{n \to \infty} \sqrt[n]{|a_n|} \). If \( L < 1 \), the series converges. If \( L > 1 \), it diverges. If \( L = 1 \), the test is inconclusive.

Step 2: Identify the Series Terms

Here, the given series is \( \sum_{n=1}^{\infty} n^{-n/2} \). Thus, the term \( a_n = n^{-n/2} \).

Step 3: Apply the Root Test Formula

We compute \( \sqrt[n]{|a_n|} = \sqrt[n]{n^{-n/2}} = (n^{-n/2})^{1/n} = n^{-1/2} \).

Step 4: Compute the Limit for the Root Test

Now, find the limit of \( n^{-1/2} \) as \( n \to \infty \). This evaluates to \( \lim_{n \to \infty} n^{-1/2} = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0 \).

Step 5: Conclude Based on the Limit

Since the limit \( L = 0 \) and \( L < 1 \), the Root Test concludes that the series \( \sum_{n=1}^{\infty} n^{-n/2} \) converges.

Key Concepts

Convergence of Series

In mathematics, when we discuss the convergence of a series, we are interested in whether the sum of the series approaches a specific number as more and more terms are added. For an infinite series \( \sum a_n \), convergence means that as you add infinitely many terms together, the sum gets closer and closer to a certain finite number. This is important because many mathematical concepts and techniques require working with convergent series to ensure meaningful results.
When we apply the Root Test to determine convergence, we calculate the limit \( L = \lim_{n \to \infty} \sqrt[n]{|a_n|} \). If this limit is smaller than 1 \((L < 1)\), the series is said to converge. In the context of the given series \( \sum_{n=1}^{\infty} n^{-n/2} \), we found that \( L = 0 \), which is less than 1, hence indicating convergence. This suggests that the series approaches a finite value as we include more terms, and thus, is convergent.

Divergence of Series

The concept of divergence is essentially the opposite of convergence. A series diverges if its sum does not approach any finite limit as more terms are added. In mathematical terms, if \( L > 1 \) using the Root Test, the series \( \sum a_n \) diverges. This means the total does not settle down to a certain number but instead grows without bound or fluctuates indefinitely.
It's also worth mentioning that if the Root Test gives \( L = 1 \), the test is inconclusive, implying that additional methods or tests need to be utilized to determine convergence or divergence. For our specific example \( \sum_{n=1}^{\infty} n^{-n/2} \), the Root Test yielded \( L = 0 \). Since 0 is less than 1, we determined the series converges, and divergence does not apply in this instance.

Limit of a Sequence

The idea of the limit is a cornerstone in calculus and analysis, particularly when dealing with series and sequences. A sequence \( \{a_n\} \) has a limit if its values get closer and closer to a certain number as \( n \) increases indefinitely. You can think of it as a laser honing in closer to a target but never perfectly touching it.

For the Root Test, understanding the limit of a sequence is crucial when determining the behavior of the root sequence \( \sqrt[n]{|a_n|} \). In our series \( \sum_{n=1}^{\infty} n^{-n/2} \), we determined the limit of the sequence \( n^{-1/2} \) is 0 as \( n \to \infty \). This limit tells us that the terms of this sequence become exceedingly small, hinting at a convergence of the series itself. Calculating such limits can serve as a powerful tool in differentiating between convergent and divergent series.