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Chapter 1: Problem 9

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. $$\sum_{n=1}^{\infty} \frac{n^{n}}{n !}$$

Short Answer

Expert verified
The series diverges by the Ratio Test.

Step by step solution

01

Apply the Ratio Test

For the given series \(\sum_{n=1}^{\infty} \frac{n^{n}}{n !} \), use the Ratio Test to determine convergence or divergence. The Ratio Test involves finding the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \(a_n = \frac{n^n}{n!} \).
02

Compute \(a_{n+1} \)

First, find \(a_{n+1} \): \(a_{n+1} = \frac{(n+1)^{n+1}}{(n+1)!} \).
03

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

Now, substitute \(a_n \) and \(a_{n+1} \) into the ratio: \(\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1)!} \times \frac{n!}{n^n} = \frac{(n+1)^{n+1}}{(n+1) n^n} \).
04

Simplify the Ratio

Simplify the expression for the limit: \(\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1) n^n} = \frac{(n+1)^n (n+1)}{(n+1) n^n} = \left( \frac{n+1}{n} \right)^n \rightarrow e \) as \(n \to \infty \). Thus, \(L = e \).
05

Convergence or Divergence

The Ratio Test states that if \(L > 1 \), the series diverges, if \(L < 1 \), the series converges, and if \(L = 1 \), the test is inconclusive. Since \(e > 1 \), the series \(\sum_{n=1}^{\infty} \frac{n^n}{n!} \) diverges.

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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 helpful tool to determine if a series converges or diverges. To use it, you calculate the limit of the absolute value of the ratio of consecutive terms: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]If this limit \( L \) is less than 1, the series converges. If \( L \) is greater than 1, the series diverges. If \( L = 1 \), the test is inconclusive.

In our example, we have the series \( \sum_{n=1}^{\infty} \frac{n^n}{n!} \). By applying the Ratio Test, we find \( L \) to be `e`, which is about 2.718, thus it is greater than 1, indicating the series diverges.
Divergence
Divergence means that the series does not sum to a finite value. In other words, the series grows without bound as more terms are added. The Ratio Test helps in identifying divergence by providing a straightforward comparison.

In our given series, \( \sum_{n=1}^{\infty} \frac{n^n}{n!} \), when \( L \) (from the Ratio Test) is greater than 1, it confirms that the series diverges. This result tells us that adding more terms will keep increasing the sum indefinitely.
Convergence
Convergence happens when the sum of the series approaches a specific value. If our Ratio Test yields a limit \( L \) less than 1, we can confidently say the series converges.

However, in our example, for the series \( \sum_{n=1}^{\infty} \frac{n^n}{n!} \), the limit was \( e \), which is greater than 1. Hence, the series does not converge. If our limit were smaller than 1, we could conclude the series converges, summing to a particular finite value.
Series
A series is the sum of the terms of a sequence. For example, in arithmetic sequences or geometric sequences, the terms follow a specific pattern. Series can either converge (sum to a finite value) or diverge (sum to infinity or oscillate without approaching a specific value). \( \sum_{n=1}^{\infty} \frac{n^n}{n!} \) is a power series, where we are dealing with factorial and exponential growth. The properties and tests, like the Ratio Test, help predict the behavior of such series without explicitly summing all terms.

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Most popular questions from this chapter

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{x}{5}\right)^{n}$$

A tall tower of circular cross section is reinforced by horizontal circular disks (like large coins), one meter apart and of negligible thickness. The radius of the disk at height \(n\) is \(1 /(n \ln n)(n \geq 2)\) Assuming that the tower is of infinite height: (a) Will the total area of the disks be finite or not? Hint: Can you compare the series with a simpler one? (b) If the disks are strengthened by wires going around their circumferences like tires, will the total length of wire required be finite or not? (c) Explain why there is not a contradiction between your answers in (a) and (b). That is, how is it possible to start with a set of disks of finite area, remove a little strip around the circumference of each, and get an infinite total length of these strips? Hint: Think about units-you can't compare area and length. Consider two cases: (1) Make the width of each strip equal to one percent of the radius of the disk from which you cut it. Now the total length is infinite but what about the total area? (2) Try to make the strips all the same width; what happens? Also see Chapter 5, Problem 3.31(b).

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{3^{2 n}}{2^{3 n}}$$

Show that \(\sum_{n=2}^{\infty} 1 / n^{3 / 2}\) is convergent. What is wrong with the following "proof" that it diverges? $$\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{27}}+\frac{1}{\sqrt{64}}+\frac{1}{\sqrt{125}}+\cdots>\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{81}}+\frac{1}{\sqrt{144}}+\cdots$$ which is $$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\dots=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right)$$. since the harmonic series diverges, the original series diverges. Hint: Compare \(3 n\) and \(n \sqrt{n}\).

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 ). $$\sum_{n=1}^{\infty} \frac{n}{n^{2}+4}$$
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