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Chapter 4: Problem 23

Use the limit comparison test to determine whether each of the following series converges or diverges. $$ \sum_{n=1}^{\infty} \frac{1}{2^{1+1 / n} n^{1+1 / n}} $$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Identify the Dominant Terms

To apply the limit comparison test, we first need to identify a series to compare with the given series. Let's simplify the expression \( \frac{1}{2^{1+1/n} n^{1+1/n}} \) by considering the dominant behavior: \( 2^{1+1/n} \approx 2 \) and \( n^{1+1/n} \approx n \). This suggests comparing the given series with the series \( \sum_{n=1}^{\infty} \frac{1}{2n} \).
02

Set Up the Limit Comparison Test

The limit comparison test involves calculating: \[ L = \lim_{n \to \infty} \frac{a_n}{b_n} \], where \( a_n = \frac{1}{2^{1+1/n} n^{1+1/n}} \) (the series term) and \( b_n = \frac{1}{2n} \) (comparison series term). We expect \( L \) to be a finite and positive number if the comparison is valid.
03

Simplify the Limit Expression

Calculate \( \frac{a_n}{b_n} = \frac{\frac{1}{2^{1+1/n} n^{1+1/n}}}{\frac{1}{2n}} = \frac{2n}{2^{1+1/n} n^{1+1/n}} = \frac{n}{2^{1/n} n^{1/n} n} = \frac{1}{2^{1/n} n^{1/n}} \). Simplifying further gives \( \frac{1}{2^{1/n} n^{1/n}} = \left(\frac{1}{2}\right)^{1/n} \cdot \left(\frac{1}{n}\right)^{1/n} \).
04

Evaluate the Limit

Take the limit:\[ L = \lim_{n \to \infty} \left(\frac{1}{2}\right)^{1/n} \cdot \left(\frac{1}{n}\right)^{1/n} \]. As \( n \to \infty \), \( \left(\frac{1}{2}\right)^{1/n} \to 1 \) and \( \left(\frac{1}{n}\right)^{1/n} \to 1 \). Hence, \[ L = 1 \cdot 1 = 1 \].
05

Conclude with the Limit Comparison Test

Since \( L = 1 \), which is a finite positive number, the limit comparison test tells us that the given series \( \sum_{n=1}^{\infty} \frac{1}{2^{1+1/n} n^{1+1/n}} \) has the same convergence behavior as \( \sum_{n=1}^{\infty} \frac{1}{2n} \). The series \( \sum_{n=1}^{\infty} \frac{1}{2n} \) is known to be a harmonic series, which diverges. Therefore, the given series also diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Convergence
When we talk about the convergence of a series, we are essentially discussing if the sum of an infinite sequence of numbers has a finite value. If the series approaches a certain limit as more terms are added, it converges. If it doesn鈥檛, we say it diverges.
Checking for convergence is crucial as it helps us understand the behavior of sequences in mathematical analysis. There are various tests to determine convergence.
One of these is the limit comparison test which compares the series in question to another series with known behavior. If it shares the same behavior, we can conclude about the convergence or divergence of the original series.
Harmonic Series
The harmonic series is a famous example in calculus characterized by the sum of reciprocals: \[ \sum_{n=1}^\infty \frac{1}{n}\].While each subsequent term gets smaller and smaller, the harmonic series actually diverges, meaning it doesn鈥檛 settle to a finite sum as more terms are added.
Even though it grows very slowly, it eventually surpasses any assigned number no matter how high. This surprising property makes it a useful benchmark for comparison tests, including the limit comparison test. In mathematical analysis, understanding the behavior of the harmonic series helps in recognizing the behavior of other intricate and complex series.
Dominant Terms
Dominant terms refer to the parts of a mathematical expression or equation that have the greatest effect on its behavior, especially as it approaches infinity. Identifying dominant terms is often a key step in simplifying complex series or functions.
When performing a limit comparison test, we estimate the dominant behavior of a series term. This involves approximating terms that change insignificantly as compared to others when the index grows very large.
In the given example, terms like \(2^{1+1/n}\) simplify to \(2\) and \(n^{1+1/n}\) to \(n\) for very large values of \(n\), showcasing how dominant terms simplify our series comparison process.
Limit Evaluation
Evaluating limits is a fundamental aspect of calculus, where we figure out what a function or sequence approaches as its input nears a certain value. In the limit comparison test, calculating limits helps us determine how two series are related.
For example, after identifying dominant terms, we sketch out the arrangement of the series-to-be-compared using the limit \[ L = \lim_{n \to \infty} \frac{a_n}{b_n} \]. Calculating this limit allows us to decide whether the two series have similar behavior based on whether \(L\) is positive and finite.
Correctly evaluating this limit showcases our comparison's validity and helps endow us with conclusions about convergence or divergence.

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Most popular questions from this chapter

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges. Let \(a_{n}=\frac{\pi^{\ln n}}{(\ln n)^{n}} .\) Show that \(\frac{a_{2 n}}{a_{n}} \rightarrow 0\) as \(n \rightarrow \infty\).

In the following exercises, use an appropriate test to determine whether the series converges. $$ a_{n}=\left(1+\frac{1}{n^{2}}\right)^{n}\left(\text { Hint }\left(1+\frac{1}{n^{2}}\right)^{n^{2}} \approx e .\right) $$

A legend from India tells that a mathematician invented chess for a king. The king enjoyed the game so much he allowed the mathematician to demand any payment. The mathematician asked for one grain of rice for the first square on the chessboard, two grains of rice for the second square on the chessboard, and so on. Find an exact expression for the total payment (in grains of rice) requested by the mathematician. Assuming there are 30,000 grains of rice in 1 pound, and 2000 pounds in 1 ton, how many tons of rice did the mathematician attempt to receive?

Does there exist a number \(p\) such that \(\sum_{n=1}^{\infty} \frac{2^{n}}{n^{p}}\) converges?

Is the series convergent or divergent? If convergent, is it absolutely convergent? $$ \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{3^{n}} $$
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