/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 Which of the sequences \(\left\\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$a_{n}=\frac{n !}{2^{n} \cdot 3^{n}}$$

Short Answer

Expert verified
The sequence diverges and does not have a limit.

Step by step solution

01

Analyze the Sequence

Consider the given sequence \(a_n = \frac{n!}{2^n \cdot 3^n}\). To determine its behavior as \(n\) approaches infinity, note that the factorial \(n!\) grows very quickly compared to exponential functions. However, \(2^n \cdot 3^n\) can be rewritten as \(6^n\), and this is an exponential function which also grows fast.
02

Examine the Growth Rates

Given \(a_n = \frac{n!}{6^n}\), note that for larger values of \(n\), \(n!\) initially grows faster than \(6^n\), but the exponential base in \(6^n\) will eventually dominate. To clarify this dominance, apply the ratio test.
03

Apply the Ratio Test

The ratio test is used to determine convergence by observing \[ L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} \frac{(n+1)!}{6^{n+1}} \cdot \frac{6^n}{n!} = \lim_{{n \to \infty}} \frac{n+1}{6} = \frac{1}{6} \cdot n + \frac{1}{6}. \] Since \(\lim_{{n \to \infty}} \frac{n+1}{6} \to \infty\), this expression will result in \(\infty\), indicating the sequence diverges.
04

Make a Conclusion

The ratio test shows that the sequence \(a_n\) grows unbounded because the limit of the ratio \( \frac{n+1}{6} \) diverges. Hence, the given sequence \(a_n = \frac{n!}{2^n \cdot 3^n}\) does not have a finite limit and diverges.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Sequence Behavior
When analyzing the behavior of a sequence, we're trying to understand what happens to its terms as we move to infinity, that is, as the index of the sequence gets very large. This can tell us whether the sequence converges to a particular value or diverges, meaning it continues to get larger or smaller without approaching any specific value.

A sequence is said to converge when its terms approach a specific number, known as the limit, as the number of terms goes to infinity. Conversely, a sequence diverges if it doesn't reach a particular limit.
  • If the terms of a sequence settle on a single value as they progress, the sequence is convergent.
  • If the terms continue to grow either positively or negatively without bound, the sequence is divergent.
In the context of the given sequence \(a_n = \frac{n!}{6^n}\), examining its behavior involves analyzing how fast the numerator and the denominator grow as \(n\) increases.
Exploring Factorial Growth
Factorial growth is one of the most rapid kinds of growth you'll encounter in sequences and functions. The notation \(n!\) represents the product of all positive integers up to \(n\), an operation that results in very large numbers quite quickly.

To illustrate:
  • \(1! = 1\)
  • \(2! = 2\times1 = 2\)
  • \(3! = 3\times2\times1 = 6\)
  • \(4! = 4\times3\times2\times1 = 24\)
  • And so on.
At higher values of \(n\), \(n!\) becomes exceedingly larger as compared to standard polynomial growth. This rapid increase is pivotal when comparing factorial functions to exponential growth, particularly because it allows for initial competition in growth rates. However, as seen in the sequence \(a_n\), despite the fast growth of \(n!\), it eventually cannot compete with the exponential growth of \(6^n\). This is a key point in understanding why the sequence behaves the way it does.
Applying the Ratio Test
The ratio test is a powerful tool for analyzing the convergence or divergence of sequences and series. It helps us evaluate the limits of sequences where ratios of successive terms can illustrate unbounded behavior.

For a sequence \(a_n\), the ratio test involves:
  • Taking the ratio of successive terms: \(\frac{a_{n+1}}{a_n}\).
  • Finding the limit of this ratio as \(n\) approaches infinity: \(L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right|\).
In the given sequence \(a_n = \frac{n!}{6^n}\), the ratio \(\frac{(n+1)!}{6^{n+1}} \cdot \frac{6^n}{n!}\) simplifies to \(\frac{n+1}{6}\). As \(n\) becomes very large, \(\frac{n+1}{6}\) tends to infinity, indicating a divergence. Thus, by applying the ratio test, we conclude succinctly that the sequence diverges. This helps validate the initial observations about the exponential domination over the factorial growth in the denominator.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Assume that each sequence converges and find its limit. $$a_{1}=3, \quad a_{n+1}=12-\sqrt{a_{n}}$$

Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: a. For what values of \(x\) can the function be replaced by each approximation with an error less than \(10^{-2} ?\). b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step \(I:\) Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(x=0\) Step 3: Calculate the \((n+1)\) st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M .\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=(1+x)^{3 / 2}, \quad-\frac{1}{2} \leq x \leq 2$$

Use the definition of convergence to prove the given limit. $$\lim _{n \rightarrow \infty}\left(1-\frac{1}{n^{2}}\right)=1$$

Assume that each sequence converges and find its limit. $$a_{1}=5, \quad a_{n+1}=\sqrt{5 a_{n}}$$

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$a_{n}=\frac{\ln n}{n}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.