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Chapter 1: Problem 5

In the following problems, find the limit of the given sequence as \(n \rightarrow \infty\). $$\frac{10^{n}}{n !}$$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Understand the Problem

We need to find the limit of the sequence as n approaches infinity for the expression \(\frac{10^n}{n!}\). This involves determining what happens to \(\frac{10^n}{n!}\) as n becomes very large.
02

Use the Ratio Test

To analyze the behavior of the sequence, we can use the ratio test. We consider the ratio of consecutive terms for large \(n\): \[\frac{a_{n+1}}{a_n} = \frac{10^{n+1}}{(n+1)!} \times \frac{n!}{10^n} = \frac{10 \cdot 10^n \cdot n!}{10^n \cdot (n+1) \cdot n!} = \frac{10}{n+1}\].
03

Analyze the Ratio

We need to check the limit of the ratio as \(n\) approaches infinity: \[\lim_{n \to \infty} \frac{10}{n+1} = 0\]. If the limit of the ratio is less than 1, then the sequence approaches 0.
04

Conclude the Limit

Since \(\lim_{n \to \infty} \frac{10}{n+1} = 0\), it indicates that the sequence \(\frac{10^n}{n!}\) converges to 0 as \(n\) approaches infinity.

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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 powerful tool used to determine the convergence or divergence of sequences and series. It examines the behavior of the ratio of consecutive terms in a sequence as the index goes to infinity. In the original problem, we applied the ratio test to the sequence \( \frac{10^n}{n!} \). Here's how it works:
  • Consider the ratio of consecutive terms: \[ \frac{a_{n+1}}{a_n} = \frac{10^{n+1}}{(n+1)!} \times \frac{n!}{10^n} = \frac{10}{n+1} \]
  • Next, take the limit of this ratio as \( n \to \infty \): \[ \frac{10}{n+1} \to 0 \]
Since 0 is less than 1, the ratio test tells us that the sequence converges to 0.
Factorial
A factorial, denoted as \( n! \), is the product of all positive integers up to \ n \. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). In our sequence, the term \( n! \) (in the denominator) grows much faster than \( 10^n \) (in the numerator) as \( n \) becomes very large. This rapid growth causes the fraction \( \frac{10^n}{n!} \) to shrink towards zero. Understanding how fast factorials grow helps explain why limits involving factorial terms often result in finite, or even zero, limits.
Convergence
Convergence refers to the behavior of a sequence or series as its terms approach a specific value. For sequences, we say a sequence \( a_n \) converges to a limit \( L \) if, as \( n \) approaches infinity, the terms \( a_n \) get arbitrarily close to \( L \). In the given sequence \( \frac{10^n}{n!} \), using the ratio test, we showed that the terms approach 0.
  • This means the sequence converges to 0.
  • The limiting behavior indicates that the terms of the sequence become negligibly small as \( n \) increases.
Such an understanding is crucial for solving many mathematical problems involving infinite processes or limits.

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Most popular questions from this chapter

Show that the Maclaurin series for \((1+x)^{p}\) converges to \((1+x)^{p}\) when \(0

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. \(\sum_{0}^{\infty} 8^{-n}\left(x^{2}-1\right)^{n}\) Method: Let \(y=x^{2}-1 .\) The power series \(\sum_{0}^{\infty} 8^{-n} y^{n}\) converges for \(|y|<8,\) so the original series converges for \(\left|x^{2}-1\right|<8,\) which means \(|x|<3\)

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{2^{3}}-\frac{1}{3^{3}}+\frac{1}{2^{4}}-\frac{1}{3^{4}}+\cdots$$

(a) Show that it is possible to stack a pile of identical books so that the top book is as far as you like to the right of the bottom book. Start at the top and each time place the pile already completed on top of another book so that the pile is just at the point of tipping. (In practice, of course, you can't let them overhang quite this much without having the stack topple. Try it with a deck of cards.) Find the distance from the right-hand end of each book to the right-hand end of the one beneath it. To find a general formula for this distance, consider the three forces acting on book \(n,\) and write the equation for the torque about its right-hand end. Show that the sum of these setbacks is a divergent series (proportional to the harmonic series). [See "Leaning Tower of The Physical Reviews," Am. J. Phys. 27, 121-122 (1959).] (b) By computer, find the sum of \(N\) terms of the harmonic series with \(N=25\) \(100,200,1000,10^{6}, 10^{100}\). (c) From the diagram in (a), you can see that with 5 books (count down from the top) the top book is completely to the right of the bottom book, that is, the overhang is slightly over one book. Use your series in (a) to verify this. Then using parts (a) and (b) and a computer as needed, find the number of books needed for an overhang of 2 books, 3 books, 10 books, 100 books.

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\ln \left(2-e^{-x}\right)$$
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