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For each of the following series, use the ratio test to determine whether the series converges or diverges. a. \(\sum_{n=1}^{\infty} \frac{2^{n^{n}}}{n !}\) b. \(\sum_{n=1}^{\infty} \frac{n^{n}}{n !} \sum_{n=1}^{\infty} \frac{(-1)^{n}(n !)^{2}}{(2 n) !}\) c. \(\sum_{n=1}^{\infty} \frac{(-1)^{n}(n !)^{2}}{(2 n) !}\)

Short Answer

Expert verified
Series a diverges, series b diverges, and series c converges.

Step by step solution

01

Understanding the Ratio Test

The ratio test can be used to determine the convergence of an infinite series \( \sum_{n=1}^{\infty} a_n \). For this test, we need to calculate the limit \( L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \). If \( L < 1 \), the series converges absolutely; if \( L > 1 \) (or \( L \) is infinite), the series diverges; if \( L = 1 \), the test is inconclusive.
02

Applying Ratio Test to Series a

For the series \( \sum_{n=1}^{\infty} \frac{2^{n^{n}}}{n !} \), let \( a_n = \frac{2^{n^{n}}}{n!} \). Calculate the ratio: \[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{2^{(n+1)^{n+1}}}{(n+1)!} \cdot \frac{n!}{2^{n^n}} \]. This simplifies roughly to \( \left( \frac{2^{n!}}{n+1} \right) \), which grows very large as \( n \to \infty \), leading \( L \to \infty \). So, the series diverges.
03

Applying Ratio Test to Series b

Here, consider the series \( \sum_{n=1}^{\infty} \frac{n^{n}}{n!} \). Let's focus on the terms effecting the convergence, \( a_n = \frac{n^n}{n!} \). Compute: \[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^{n+1}}{(n+1)!} \cdot \frac{n!}{n^n} \], simplifying to \( \left( \frac{(n+1)^n}{n^n} \cdot \frac{1}{n+1} \right) \), which leads \( L \) to be greater than 1. Thus, this series diverges.
04

Applying Ratio Test to Series c

For the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(n !)^{2}}{(2 n) !} \), focus on the absolute value \( a_n = \frac{(n!)^2}{(2n)!} \). Consider the ratio: \[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{((n+1)!)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2}\], and as \( n \to \infty \), \( L \to 0 \) since the factorial growth in the denominator dominates the numerator. Therefore, this series converges absolutely.

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Key Concepts

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

Series Convergence
Series Convergence is an important concept that helps us understand whether a series adds up to a finite value or not. There are different ways to test for convergence, and the ratio test is one such method. When we apply the ratio test, we focus on the terms of the series as they progress to infinity.
  • If the limit of the ratio of consecutive terms is less than 1, the series converges absolutely.
  • If the limit is greater than 1 or infinite, the series diverges.
  • If the limit equals 1, the test is inconclusive, and other methods might be needed to assess convergence.
Understanding convergence helps predict the behavior of series and functions in various mathematical applications, which is crucial in areas such as calculus and mathematical analysis.
Factorial Growth
Factorial Growth describes how quickly the factorial of a number increases as the number itself increases. The notation \(n!\) is used to denote the factorial, which is the product of all positive integers up to \(n\). This growth is extremely rapid because each new term in the sequence is the previous term multiplied by a larger number.
  • Even for relatively small values of \(n\), \(n!\) grows faster than exponential functions like \(2^n\) or \(n^2\).
  • This dramatic growth can heavily influence the behavior of a series, as observed in the original exercise, where the denominator \((n!)\) significantly affects convergence.
When using the ratio test or examining series, understanding factorial growth becomes important because it can determine whether terms shrink fast enough as \(n\) approaches infinity to ensure convergence.
Infinite Series
An Infinite Series is a sum of infinite terms. Imagine continuously adding a set of numbers, one after another, forever. The sum of an infinite series is given by the notation \(\sum_{n=1}^{\infty} a_n\), where \(a_n\) represents each term in the series. The concepts of convergence and divergence determine whether this infinite process yields a finite sum. Mathematicians have devised various tests, like the ratio test used in the original solutions, to study these series.
  • Understanding infinite series is essential in calculus, where they often represent functions as power series, like Taylor or Maclaurin series.
  • Despite the infinite number of terms, some series converge (add up to a specific value), while others diverge (grow indefinitely).
Grasping the idea of infinite series is key to solving real-world problems where approximation and summation are needed, such as physics and engineering applications.

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Most popular questions from this chapter

Does there exist a number \(p\) such that \(\sum_{n=1}^{\infty} \frac{2^{n}}{n^{p}}\) converges?

The following problems consider a simple population model of the housefly, which can be exhibited by the recursive formula \(x_{n+1}=b x_{n}\), where \(x_{n}\) is the population of houseflies at generation \(n\), and \(b\) is the average number of offspring per housefly who survive to the next generation. Assume a starting population \(x_{0}\). For what values of \(b\) will the series converge and diverge? What does the series converge to?

Is the series convergent or divergent? If convergent, is it absolutely convergent? $$ \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{3^{n}} $$

The following alternating series converge to given multiples of \(\pi .\) Find the value of \(N\) predicted by the remainder estimate such that the Nth partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum \(N\) for which the error bound holds, and give the desired approximate value in each case. Up to 15 decimals places, \(\pi=3.141592653589793 .\)[T] The Euler transform rewrites \(S=\sum_{n=0}^{\infty}(-1)^{n} b_{n}\) as \(S=\sum_{n=0}^{\infty}(-1)^{n} 2^{-n-1} \sum_{m=0}^{n}\left(\begin{array}{c}n \\ m\end{array}\right) b_{n-m}\). For the alternating harmonic series, it takes the form \(\ln (2)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=\sum_{n=1}^{\infty} \frac{1}{n 2^{n}} .\) Compute partial sums of \(\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\) until they approximate \(\ln (2)\) accurate to within \(0.0001\). How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate \(\ln (2)\).

Is the series convergent or divergent? If convergent, is it absolutely convergent? $$ \sum_{n=1}^{\infty} \cos (\pi n) e^{-n} $$

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