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Chapter 9: Problem 7

Use the Ratio Test to determine convergence or divergence. $$ \sum_{n=1}^{\infty} \frac{n !}{n^{100}} $$

Short Answer

Expert verified
The series converges by the Ratio Test.

Step by step solution

01

Identify General Term

The general term of the series is given as \(a_n = \frac{n!}{n^{100}}\).
02

Apply the Ratio Test Formula

The Ratio Test states that a series \( \sum a_n \) converges if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). We need to compute this limit for our series.
03

Write the Expression for \(a_{n+1}\)

We start by expressing \(a_{n+1}\): \(a_{n+1} = \frac{(n+1)!}{(n+1)^{100}}\).
04

Set Up Ratio \(\frac{a_{n+1}}{a_n}\)

Now, substitute \(a_n\) and \(a_{n+1}\) into the ratio: \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{100}} \cdot \frac{n^{100}}{n!} = \frac{n^{100} \cdot (n+1)}{(n+1)^{100} \cdot n!} \].
05

Simplify the Ratio

The expression simplifies to \[ \frac{a_{n+1}}{a_n} = \frac{n^{100} \cdot (n+1)}{n! \cdot (n+1)^{100}} = \frac{n^{100}}{(n+1)^{99}} \cdot \frac{1}{n!} \].
06

Evaluate the Limit

Calculate the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n^{100}}{(n+1)^{99}} \cdot \frac{1}{n!}\). As \(n\) grows large, \(n!\) increases much faster than \(n^{100}\), making the limit equal to 0.
07

Conclude with the Ratio Test Result

Since \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1\), the Ratio Test tells us that the series \( \sum_{n=1}^{\infty} \frac{n!}{n^{100}} \) converges.

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Key Concepts

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

Convergence
Convergence is a crucial concept when studying series in mathematics. It refers to whether a series approaches a specific value as you add more terms. For our series, we need to determine if the sequence of partial sums \[ S_n = a_1 + a_2 + ... + a_n \] approaches a finite limit as \( n \) goes to infinity.
The Ratio Test is a popular tool to find convergence. It involves checking the limit of the ratio of consecutive terms. If this limit is less than one, the series converges.
In the problem, we apply the Ratio Test to decide on the convergence of \( \sum_{n=1}^{\infty} \frac{n!}{n^{100}} \). The ratio simplifies to something that tends towards zero, thus confirming convergence.
Factorial
A factorial, denoted as \( n! \), is the product of all positive integers up to \( n \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). This operation grows very rapidly compared to simple exponentiation.
  • Factorials are used extensively in permutations, combinations, and series.
  • In the series given, the factorial term \( n! \) in the numerator dictates the growth of the terms as \( n \) increases.
During the evaluation, we saw that \( n! \) grows much faster than \( n^{100} \), which is critical in determining the convergence through the Ratio Test.
Infinite Series
Infinite series are sums with an infinite number of terms. It's like adding an endless list of numbers. The sum can either reach a specific value (converge) or not (diverge). Understanding infinite series is essential to solving many problems in calculus and complex analysis.
To solve the exercise, we investigated the series \( \sum_{n=1}^{\infty} \frac{n!}{n^{100}} \). The terms decrease according to their position in the sequence, which simplifies the computation when using the Ratio Test.
Mathematical Limits
Limits help us understand behavior as variables approach specific values. They are foundational to calculus and analysis, assisting in defining derivatives, integrals, and series behavior.
  • In this context, we analyzed the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • This limit is key to applying the Ratio Test effectively.
The solution showed that even for large values of \( n \), the ratio of consecutive terms tends towards zero, proving the series converges. The concept of limits is critical for exploring such patterns in mathematics.

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Most popular questions from this chapter

Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems \(43-48\) to see that you get the same answers using the methods of Section \(9.7 .\) \(3 \sin x-2 \exp x\)

One can sometimes find a Maclaurin series by the method of equating coefficients. For example, let $$ \tan x=\frac{\sin x}{\cos x}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots $$ Then multiply by \(\cos x\) and replace \(\sin x\) and \(\cos x\) by their series to obtain $$ \begin{aligned} x-\frac{x^{3}}{6}+\cdots &=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots\right)\left(1-\frac{x^{2}}{2}+\cdots\right) \\ &=a_{0}+a_{1} x+\left(a_{2}-\frac{a_{0}}{2}\right) x^{2}+\left(a_{3}-\frac{a_{1}}{2}\right) x^{3}+\cdots \end{aligned} $$ Thus, $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}-\frac{a_{0}}{2}=0, \quad a_{3}-\frac{a_{1}}{2}=-\frac{1}{6}, \quad \cdots $$ so $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}=0, \quad a_{3}=\frac{1}{3}, \quad \cdots $$ and therefore $$ \tan x=0+x+0+\frac{1}{3} x^{3}+\cdots $$ which agrees with Problem 1 . Use this method to find the terms through \(x^{4}\) in the series for \(\sec x\).

Let $$ f(t)=\left\\{\begin{array}{ll} 0 & \text { if } t<0 \\ t^{4} & \text { if } t \geq 0 \end{array}\right. $$ Explain why \(f(t)\) cannot be represented by a Maclaurin series. Also show that, if \(g(t)\) gives the distance traveled by a car that is stationary for \(t<0\) and moving ahead for \(t \geq 0, g(t)\) cannot be represented by a Maclaurin series.

Plot on the same axes the given function along with the Maclaurin polynomials of orders \(1,2,3\), and \(4 .\) $$ \cos 2 x $$

Determine convergence or divergence for each of the series. Indicate the test you use. $$ \sum_{n=2}^{\infty}\left(1-\frac{1}{n}\right)^{n} $$
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