Question

Use the limit comparison test to determine whether each of the following series converges or diverges. $$ \sum_{n=1}^{\infty}\left(\frac{\ln n}{n^{0.6}}\right)^{2} $$

Short answer

The series diverges.

Step-by-step solution

Identify the terms of the series

We have the series \( \sum_{n=1}^{\infty} \left( \frac{\ln n}{n^{0.6}} \right)^{2} \). The terms of this series are \( a_n = \left( \frac{\ln n}{n^{0.6}} \right)^{2} = \frac{(\ln n)^2}{n^{1.2}} \). We need to compare this with a known convergent or divergent series for the limit comparison test.

Choose a comparison series

For the limit comparison test, choose a series \( \sum b_n \) where \( b_n = \frac{1}{n^{1.2}} \). This is a p-series with \( p=1.2 \). We know that a p-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \). Hence, \( \sum \frac{1}{n^{1.2}} \) converges.

Apply the limit comparison test

Compute the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{(\ln n)^2}{n^{1.2}}}{\frac{1}{n^{1.2}}} = \lim_{n \to \infty} (\ln n)^2 \).

Evaluate the limit

We note that \( (\ln n)^2 \to \infty \) as \( n \to \infty \). Therefore, the limit \( L \) does not exist in a way that allows us to apply the limit comparison test directly since it is not a finite positive number.

Conclusion

Since the limit \( L \) is infinite, which means \( a_n \sim k b_n \) with \( k \to \infty \), this suggests \( a_n \) is larger than \( b_n \) for large \( n \). However, this does not provide the needed information to use the limit comparison test effectively. We instead use the comparison test to note that because \( (\ln n)^2 \) grows without bounds, \( a_n \) cannot be smaller than some convergent series after some point. Thus, the original series diverges.

Key concepts

Series Convergence

Understanding series convergence is an essential concept in calculus and mathematical analysis. When we talk about the convergence of a series, we're exploring whether the sum of its infinite terms reaches a finite limit. This topic can seem a bit tricky at first, but it's all about whether we end up with a finite number or not.

When a series doesn't converge, we call it divergent. In simple terms, if the terms add up to infinity or oscillate without settling to a point, the series is divergent. To deal with these concepts, mathematicians use various tests to analyze whether a given series converges or diverges.

One of the core methods for investigating series convergence is the "comparison test", which often pairs with the limit comparison test. These techniques involve comparing the given series with another known series. If the new series' behavior is known—whether it converges or diverges—we can infer the behavior of the original series. Such tools are valuable because many complicated series can be compared to simpler, well-known forms like the geometric series or the p-series.

p-Series

A p-series is a specific type of mathematical series that takes the form \( \sum \frac{1}{n^p} \), where \( n \) is a positive integer starting from 1, and \( p \) is a real number. The convergence or divergence of a p-series depends heavily on the value of \( p \).

When \( p > 1 \), the p-series converges, meaning that the sum of all its terms approaches a finite number as more and more terms are added. Conversely, when \( p \leq 1 \), the series diverges, extending indefinitely without reaching a finite limit. This property makes p-series a convenient tool for comparison because we can quickly determine the behavior of a more complex series by comparing it to a related p-series.

In practice, if we encounter a series such as \( \sum \frac{1}{n^{1.2}} \), we know it converges because it's a p-series with \( p = 1.2 > 1 \). This principle often underlies problems where the limit comparison test is used, as seen in exercises that involve breaking down series to simpler forms for easier analysis.

Limit Evaluation

Limit evaluation is crucial when applying the limit comparison test for series convergence. This test involves comparing the series of interest to a benchmark series that behaves predictably.

To apply the limit comparison test, you compute the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n \) are the terms of the original series, and \( b_n \) are the terms of the comparison series. For the test to be applicable, \( L \) must be a positive, finite number. This means the terms of both series grow or shrink at similar rates as \( n \) approaches infinity.

If this limit \( L \) exists and is positive, then both series either converge or diverge together. In contrast, if \( L \) is infinite or undefined, it indicates that the original series does not mirror the behavior of the comparison series in a manner that supports a straightforward conclusion through the limit comparison test. That's why understanding limits and how they can be used to assess the behavior of terms is integral to series convergence analysis. Limit tests rely on the balanced ratio of these terms to guide us to the convergence or divergence of the series.