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Chapter 8: Problem 64

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{3^{n}}{n !}\right\\}$$

Short Answer

Expert verified
Answer: The sequence converges.

Step by step solution

01

Write down the sequence

The given sequence is: $$\left\\{\frac{3^{n}}{n!}\right\\}$$
02

Apply Ratio Test

To apply the Ratio Test, we need to find the limit of the ratio of consecutive terms: $$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$$ Plug the expressions for \(a_{n+1}\) and \(a_n\) of the given sequence into the equation: $$\lim_{n\to\infty} \left|\frac{\frac{3^{n+1}}{(n+1)!}}{\frac{3^n}{n!}}\right|$$
03

Simplify the Ratio

Now, we need to simplify the ratio by canceling common terms: $$\lim_{n\to\infty} \left|\frac{3^{n+1}n!}{3^n(n+1)!}\right|$$ Divide both numerator and denominator by \(3^n\): $$\lim_{n\to\infty} \left|\frac{3n!}{(n+1)!}\right|$$
04

Cancel out the factorial terms

Since \((n+1)! = (n+1)n!\), we can cancel out the factorial terms in the ratio: $$\lim_{n\to\infty} \left|\frac{3n!}{(n+1)n!}\right| = \lim_{n\to\infty} \left|\frac{3}{n+1}\right|$$
05

Find the limit

Evaluate the limit of the simplified expression as n tends to infinity: $$\lim_{n\to\infty} \left|\frac{3}{n+1}\right| = 0$$ Since the limit is less than 1, by the Ratio Test, the sequence converges.

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Key Concepts

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

Limit of a sequence
In mathematics, the limit of a sequence describes the value that the terms of a sequence "approach" as the index (usually denoted as \( n \)) becomes infinitely large. To understand this, picture a sequence as an infinite list of numbers. As you progress towards infinity, if these numbers get closer and closer to a specific value, we say that the sequence has a limit.
For example, in the sequence given in the exercise, \( \left\{\frac{3^n}{n!}\right\} \), we are interested in understanding what happens as \( n \) grows larger. Using specific mathematical techniques like the Ratio Test, one can determine the behavior of a sequence at infinity. When the limit exists and leads to a particular finite number, the sequence converges to that limit.
Ratio Test
The Ratio Test is a powerful tool used in calculus to determine the convergence or divergence of sequences and series. The test revolves around examining the ratio of consecutive terms within a sequence.

To apply the Ratio Test:
  • Take the absolute value of the ratio of the \((n+1)\)-th term to the \(n\)-th term: \( \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • Simplify the expression as much as possible.
  • Calculate the limit of this simplified ratio as \( n \to \infty \).
If this limit is less than 1, the sequence converges. If it is greater than 1, the sequence diverges. If it equals 1, the test is inconclusive. In the exercise we solved, the calculation led to a limit of 0, which is less than 1, indicating that the sequence converges.
Convergence and divergence
Convergence and divergence describe whether a sequence or series approaches a specific value as it continues indefinitely. These concepts are fundamental in the analysis of sequences.

  • **Convergence:** A sequence converges when it approaches a definite limit as \( n \to \infty \). This means the terms of the sequence become arbitrarily close to a specific number.
  • **Divergence:** A sequence diverges if it fails to settle on a single number. This could mean the terms grow without bound, oscillate, or have no discernible pattern.
In our context, applying the Ratio Test showed that the given sequence converges. Understanding convergence and divergence helps predict long-term behaviors, crucial for an array of applications in science, engineering, and finance.

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Most popular questions from this chapter

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{n^{10}}{\ln ^{1000} n}\right\\}$$

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)},$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). a. Explain why \(\left\\{F_{n}\right\\}\) is a decreasing sequence. b. Plot \(\left\\{F_{n}\right\\},\) for \(n=1,2, \ldots, 20\). c. Based on your experiments, make a conjecture about \(\lim _{n \rightarrow \infty} F_{n}\).

The expression $$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+}}}}.$$ where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+1 / a_{n},\) for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of $$a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\frac{b}{a+}}}}$$ where \(a\) and \(b\) are positive real numbers.

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{e^{n / 10}}{2^{n}}\right\\}$$

Prove that if \(\left\\{a_{n}\right\\} \ll\left\\{b_{n}\right\\}\) (as used in Theorem 8.6 ), then \(\left\\{c a_{n}\right\\} \ll\left\\{d b_{n}\right\\},\) where \(c\) and \(d\) are positive real numbers.
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