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Chapter 7: Problem 37

Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{(2 n)^{n}}\)

Short Answer

Expert verified
The sequence diverges.

Step by step solution

01

Identify the nth term and the (n+1)th term

The nth term provided in the exercise is \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{(2 n)^{n}}\). To find the (n+1)th term, simply replace 'n' with 'n+1' to get \(a_{n+1}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n+1-1)}{(2 (n+1))^{n+1}} = \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot 2n \cdot 2n+1}{(2n+2)^{n+1}}\).
02

Apply the Ratio Test

Using the Ratio Test to determine the convergence or divergence of the sequence involves dividing the (n+1)th term by the nth term, and taking the limit as n tends to infinity. \(\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}} =\lim_{n\to\infty} \frac{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot 2n \cdot 2n+1}{(2n+2)^{n+1}}}{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{(2 n)^{n}}} = \lim_{n\to\infty} \frac{(2n+1)(2n)^n}{(2n+2)^{n+1}}\).
03

Simplify the expression

Simplify the expression obtained in Step 2 to render its limit calculation easier. This gives \(\lim_{n\to\infty} \frac{(2n+1)}{(2n+2)}\).
04

Calculate the limit

Calculate the limit as n tends to infinity for \(\lim_{n\to\infty} \frac{(2n+1)}{(2n+2)}\). As both numerator and denominator tend to infinity, this is an indeterminate form of type \(\frac{\infty}{\infty}\). Dividing both numerator and denominator by n yields \(\lim_{n\to\infty} \frac{2+(1/n)}{2+(2/n)}\). Taking the limit as n tends to infinity gives the ratio \(\frac{2}{2} = 1\).
05

Determine convergence or divergence

As the value of the limit (ratio) is 1, which is not less than 1, so according to the Ratio Test, the sequence diverges.

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Key Concepts

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

Ratio Test
The Ratio Test is a popular method for checking whether a series converges or diverges. It is especially helpful for series with terms that involve factorials, powers, or combinations thereof. Generally, the test involves:
  • Finding the ratio of subsequent terms in the series, typically the (n+1)th term divided by the nth term.
  • Taking the limit of this ratio as n approaches infinity.
  • Checking the result of this limit:
    • If the limit is less than 1, the series converges absolutely.
    • If the limit is greater than 1, or infinite, the series diverges.
    • If the limit equals 1, the test is inconclusive, and another method may be needed.
In the given exercise, after calculating the limit of the ratio, it equals 1. Here, the Ratio Test becomes inconclusive, suggesting a need for further investigation into the behavior of the series.
Limit Calculation
Limit calculation is a crucial skill in analyzing the convergence of sequences and series. It involves finding the value that a function or sequence approaches as the input approaches some value. In many cases, especially in sequence-related problems, the input approaches infinity.To perform a limit calculation:
  • Simplify the expression as much as possible, often by canceling terms or factoring.
  • Identify if the expression forms an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
  • If indeterminate, use tools like L'H么pital's Rule or algebraic manipulation to resolve it.
In this example, simplifying the limit for \(\lim_{n\to\infty} \frac{(2n+1)(2n)^n}{(2n+2)^{n+1}}\) helps reveal essential properties of the sequence and determine its behavior as \(n\rightarrow\infty\).
Indeterminate Forms
Indeterminate forms are expressions that do not directly point to a specific value or conclusion. They occur often in calculus and often indicate the need for further manipulation to find a limit. Common forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), and more.
  • In these situations, insights may be achieved by algebraic simplification or calculus-based theorems like L'H么pital's Rule, which involves differentiating the numerator and denominator.
  • Understanding these forms is vital in limit calculation processes.
When evaluating the limit \(\lim_{n\to\infty} \frac{(2n+1)}{(2n+2)}\), it initially presents an indeterminate form \(\frac{\infty}{\infty}\). By rewriting the expression using simple algebra, the limit can eventually be resolved to a finite number, \(1\).
Divergence of Sequences
The divergence of sequences indicates that the terms do not settle towards a single finite value as the sequence progresses to infinity. A sequence that diverges may increase infinitely, decrease without bound, or oscillate between values.Key indicators that a sequence diverges include:
  • The terms increase or decrease without approaching a limit.
  • Applying tests, like the Ratio Test, provide limits \(> 1\) or otherwise inconclusive results, such as equalling 1, which suggests non-convergence.
In this context, the exercise ends with an inconclusive Ratio Test where the limit equals 1, pointing to the divergence of the sequence as it lacks convergence to a specific limit.

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Most popular questions from this chapter

Determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty}(1.075)^{n} $$

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right) $$

Use a graphing utility to graph the function. Identify the horizontal asymptote of the graph and determine its relationship to the sum of the series. $$ \frac{\text { Function }}{f(x)=2\left[\frac{1-(0.8)^{x}}{1-0.8}\right]} \frac{\text { Series }}{\sum_{n=0}^{\infty} 2\left(\frac{4}{5}\right)^{n}} $$

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(|r|<1,\) then \(\sum_{n=1}^{\infty} a r^{n}=\frac{a}{(1-r)} .\)

Consider the formula \(\frac{1}{x-1}=1+x+x^{2}+x^{3}+\cdots\) Given \(x=-1\) and \(x=2\), can you conclude that either of the following statements is true? Explain your reasoning. (a) \(\frac{1}{2}=1-1+1-1+\cdots\) (b) \(-1=1+2+4+8+\cdots\)
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