/*! 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 36 Converge, and which diverge? Giv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$ \sum_{n=1}^{\infty} \frac{n^{n}}{n !} $$

Short Answer

Expert verified
The series diverges because the limit in the Ratio Test is \( e > 1 \).

Step by step solution

01

Understand the Series

The series we are dealing with is given by \( \sum_{n=1}^{\infty} \frac{n^{n}}{n!} \). Our goal is to determine whether this series converges or diverges, and if it converges, find its sum.
02

Consider the Ratio Test

The Ratio Test is a good candidate for this series since it involves factorials, which often simplify nicely with the test. The Ratio Test states that if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), then the series \( \sum a_n \) converges.
03

Compute \( a_{n+1} \) and \( a_n \)

For the given series, each term \( a_n = \frac{n^n}{n!} \). The next term would be \( a_{n+1} = \frac{(n+1)^{n+1}}{(n+1)!} \).
04

Compute the Ratio \( \frac{a_{n+1}}{a_n} \)

Evaluate the ratio:\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1} \cdot n!}{n^n \cdot (n+1)!} = \frac{(n+1)^{n+1}}{n^n (n+1)} \]Simplifying the expression:\[ = \frac{(n+1)^n}{n^n} = \left( 1 + \frac{1}{n} \right)^n \]
05

Evaluate the Limit

Take the limit of \( \left( 1 + \frac{1}{n} \right)^n \) as \( n \to \infty \). This expression approaches \( e \) (the base of the natural logarithm), specifically:\[ \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e \]
06

Conclusion Using the Ratio Test

Since the limit calculated is \( e > 1 \), the Ratio Test indicates that the series diverges. Thus, the series \( \sum_{n=1}^{\infty} \frac{n^n}{n!} \) 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.

Ratio Test
The Ratio Test is a handy tool to determine the convergence or divergence of an infinite series. It's particularly useful when factorials or exponentiation are involved in the series terms.

To apply the Ratio Test, we evaluate \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|, \]where \( a_n \) represents the general term of the series.
  • If this limit is less than 1, the series converges.
  • If it equals to 1, the test is inconclusive.
  • If it exceeds 1, the series diverges.
The idea is that if the next term is substantially less than the current term as \( n \) becomes large, then the series terms are shrinking rapidly enough for the series to converge. Otherwise, divergence is expected.

In this exercise, we used the Ratio Test on the series \( \sum_{n=1}^{\infty} \frac{n^n}{n!} \). After computing the ratio of successive terms and simplifying, we derived an expression that approaches \( e \) as \( n \) increases. Since \( e > 1 \), the series diverges.
Factorial Function
The factorial function, denoted as \( n! \), is a fundamental concept in mathematics. It is defined as the product of an integer and all the positive integers below it. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

This function grows extremely rapidly and is often used in series to express permutation counts or in scenarios involving combinatorics.

Factorials can simplify in ratio tests because their explosive growth rates dominate other terms, like power terms in sequence discussions. In this exercise, the presence of \( n! \) in the denominator plays a crucial role when applying the Ratio Test to check for convergence of the series. Despite the presence of \( n! \), the series diverges due to the exponential growth of \( n^n \). This highlights the powerful nature of factorials in altering the behavior of sequences and series.
Exponentiation in Sequences
Exponentiation in sequences involves raising a number to the power of a variable, often \( n \) in series contexts. It signifies a rapid growth since \[ n^n = n \times n \times n \times \ldots \times n \] (a total of \( n \) times).

This growth is much quicker than linear or polynomial growth and can outpace other sequence components if not controlled by a dominant factorial.

In the sum \( \sum_{n=1}^{\infty} \frac{n^n}{n!} \), the term \( n^n \) is in the numerator. Despite the factorial in the denominator, \( n^n \) grows too quickly, leading to divergence. Understanding the behavior of sequences involving exponentiation is crucial to predict their influence on overall series behavior.

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

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) 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 \(L ?\) $$ a_{n}=\sqrt[n]{n} $$

The value of \(\sum_{n=1}^{\infty} \tan ^{-1}\left(2 / n^{2}\right)\) a. Use the formula for the tangent of the difference of two angles to show that $$ \tan \left(\tan ^{-1}(n+1)-\tan ^{-1}(n-1)\right)=\frac{2}{n^{2}} $$ b. Show that $$ \sum_{n=1}^{N} \tan ^{-1} \frac{2}{n^{2}}=\tan ^{-1}(N+1)+\tan ^{-1} N-\frac{\pi}{4} $$ c. Find the value of \(\sum_{n=1}^{\infty} \tan ^{-1} \frac{2}{n^{2}}\)

Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{e^{n}}{1+e^{2 n}} $$

Outline of the proof of the Rearrangement Theorem (Theo- rem 17\()\) a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n},\) and let \(s_{k}=\sum_{n=1}^{k} a_{n}\) . Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\) $$\sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \quad\( and \)\quad\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2}$$ since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\},\) there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3},\) then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1} .\) Therefore, if \(n \geq N_{3}\) $$\begin{aligned}\left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned}$$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\)

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+y=0, \quad y^{\prime}(0)=0 \text { and } y(0)=1 $$

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.