Question

For each of the following sequences, if the divergence test applies, either state that \(\lim _{n \rightarrow \infty} a_{n}\) does not exist or find \(\lim _{n \rightarrow \infty} a_{n} .\) If the divergence test does not apply, state why. \(a_{n}=\frac{(\ln n)^{2}}{\sqrt{n}}\)

Short answer

The divergence test does not apply as \(\lim_{n \rightarrow \infty} a_n = 0\).

Step-by-step solution

Determine Applicability of Divergence Test

The divergence test applies if \(\lim_{n \rightarrow \infty} a_{n} eq 0\). First, we need to find the limit of \(a_n = \frac{(\ln n)^2}{\sqrt{n}}\) as \(n\) approaches infinity.

Find the Limit of the Sequence

Find the limit \(\lim_{n \rightarrow \infty} \frac{(\ln n)^2}{\sqrt{n}}\). Notice that as \(n\) increases, \(\ln n\) grows slower than \(\sqrt{n}\). However, the square \((\ln n)^2\) complicates the limit calculation. We can use L'Hopital's Rule to evaluate this limit due to its indeterminate form \(\frac{\infty}{\infty}\).

Applying L'Hopital's Rule

Rewrite the sequence into a form suitable for L'Hopital's Rule: \(\lim_{n \rightarrow \infty} \frac{(\ln n)^2}{n^{1/2}}\). Rewriting as \(\lim_{n \rightarrow \infty} \frac{(\ln n)^2}{n^{1/2}}\) and differentiate the numerator and the denominator with respect to \(n\).First derivative: - Numerator: \(d/dn [(\ln n)^2] = 2 \ln n / n\)- Denominator: \(d/dn[n^{1/2}] = \frac{1}{2} n^{-1/2}\)The new limit becomes: \(\lim_{n \rightarrow \infty} \frac{2 \ln n / n}{1/2 \, n^{-1/2}} = \lim_{n \rightarrow \infty} \frac{4 \ln n}{n^{1/2}}\).Apply L'Hopital's Rule again using these derivatives:- Numerator: \(d/dn [4 \ln n] = \frac{4}{n}\)- Denominator: \(d/dn [n^{1/2}] = \frac{1}{2} n^{-1/2}\)The limit becomes: \(\lim_{n \rightarrow \infty} \frac{4 / n}{1/ 2 \, n^{-1/2}} = \lim_{n \rightarrow \infty} \frac{8}{n^{1/2}}\).

Evaluate the Limit

Now simplify \(\lim_{n \rightarrow \infty} \frac{8}{n^{1/2}}\): as \(n\) approaches infinity, \(n^{1/2}\) grows very large, pushing the fraction \(\frac{8}{n^{1/2}} \) towards zero. Thus, \(\lim_{n \rightarrow \infty} \frac{8}{n^{1/2}} = 0\).

Conclusion Based on Divergence Test

Since the limit \(\lim_{n \rightarrow \infty} a_n = 0\), the divergence test does not apply. The divergence test states that if \(\lim_{n \rightarrow \infty} a_n eq 0\), the series diverges, otherwise the test is inconclusive. Therefore, based on the divergence test, we can't conclude whether the series converges or diverges just from \(a_n\).

Key concepts

Limit of a sequence

In mathematical sequences, a "limit" describes the value that the terms of the sequence approach as they progress indefinitely. The limit of a sequence is fundamental for determining the behavior of the sequence as it extends to infinity.

When we say \(\to\infty\), it implies that we are observing the behavior of the sequence as it goes towards an infinitely large integer. A common technique for finding the limit involves comparing the growth rates of terms within the sequence. For example, given a sequence like \(a_n = \frac{(\ln n)^2}{\sqrt{n}}\), practitioner may analyze the behavior by observing how \(\ln n\) and \(\sqrt{n}\) grow relative to each other.

Here, \(\ln n\) grows slower than \(\sqrt{n}\), which provides information about where the limit might lie. Using rules or formulas, such as L'Hopital's Rule, can often simplify this process when the limit isn't easily visible.

L'Hopital's Rule

L'Hopital's Rule is an essential tool in calculus used for finding limits of indeterminate forms, often symbolized by expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When faced with such uncertainty, L'Hopital's Rule allows us to differentiate the numerator and the denominator separately.

Let's apply this to our sequence: \(\lim_{n \to \infty} \frac{(\ln n)^2}{\sqrt{n}}\). Initially, this gives the indeterminate \(\frac{\infty}{\infty}\) form. To resolve this, we take the derivative of the numerator \(d/dn[(\ln n)^2] = \frac{2 \ln n}{n}\) and denominator \(d/dn[n^{1/2}] = \frac{1}{2}n^{-1/2}\), which simplifies the sequence so that we can evaluate it using L'Hopital's Rule again.

Upon applying the rule a second time, the expression reduces further, simplifying the limit problem until a direct conclusion about the sequence's behavior can be reached.

Series convergence and divergence

The concept of convergence and divergence in the context of series is fundamental in understanding how sum behaves when all its terms are added up infinitely. If a series converges, the sum of its infinite terms approaches a finite limit. If it diverges, it may grow towards infinity or behave erratically without approaching any particular value.

The divergence test is one of several tests that help determine whether a series converges or diverges. It specifically tells us if \(\lim_{n \to \infty} a_n eq 0\), then the series must diverge. However, if the limit equals zero, the test is inconclusive and indicates that other methods must be employed to conclude about convergence.

For instance, in our sequence \(a_n = \frac{(\ln n)^2}{\sqrt{n}}\), we found that its limit as \(n \to \infty\) is zero. Since the divergence test is inconclusive in this case, more sophisticated approaches are necessary to determine if the series truly converges or diverges.