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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

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. $$\sum_{n=1}^{\infty} a_{n} \quad \text { if } a_{n+1}=\frac{n}{2 n+3} a_{n}$$

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\frac{d^{10}}{d x^{10}}\left(x^{8} \tan ^{2} x\right) \quad \text { at } x=0$$

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. $$\sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{10^{n}}$$

(a) It is clear that you (or your computer) can't find the sum of an infinite series just by adding up the terms one by one. For example, to get \(\zeta(1.1)=\) \(\sum_{n=1}^{\infty} 1 / n^{1.1}(\text { see Problem } 15.22)\) with error \(<0.005\) takes about \(10^{33}\) terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed \(N\) terms, the sum \(R_{N}\) of the rest of the series is between \(I_{N}=\int_{N}^{\infty} a_{n} d n\) and \(I_{N+1}=\int_{N+1}^{\infty} a_{n} d n\) (b) Find the integrals in (a) for the \(\zeta(1.1)\) series and verify the claimed number of terms needed for error \(<0.005 .\) Hint: Find \(N\) such that \(I_{N}=0.005 .\) Also find upper and lower bounds for \(\zeta(1.1)\) by computing \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N}^{\infty} n^{-1.1} d n\) and \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N+1}^{\infty} n^{-1.1} d n\) where \(N\) is far less than \(10^{33} .\) Hint: You want the difference between the upper and lower limits to be about \(0.005 ;\) find \(\mathrm{N}\) so that term \(a_{N}=0.005\).

Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals (see Problem 16 ). $$\sum_{n=1}^{\infty} \frac{e^{n}}{e^{2 n}+9}$$
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