/*! 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 33 Which of the series in Exercises... [FREE SOLUTION] | 91Ó°ÊÓ

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Which of the series in Exercises \(15 - 48\) converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum _ { n = 1 } ^ { \infty } \frac { ( - 100 ) ^ { n } } { n ! } $$

Short Answer

Expert verified
The series converges absolutely by the ratio test.

Step by step solution

01

Identify the Series

We are given the series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \). This is an alternating series because of the factor \((-100)^n\), which alternates the sign of each term.
02

Check Absolute Convergence

To check for absolute convergence, consider the series of absolute values, \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \). We will use the ratio test to determine if this series converges.
03

Apply the Ratio Test

The ratio test involves looking at the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). For our series, we have \( a_n = \frac{100^n}{n!} \) and \( a_{n+1} = \frac{100^{n+1}}{(n+1)!} \). The ratio becomes:\[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{100^{n+1} \cdot n!}{100^n \cdot (n+1)!} = \frac{100}{n+1} \] Then take the limit: \[ \lim_{n \to \infty} \frac{100}{n+1} = 0 \]
04

Interpret the Ratio Test Result

Since the limit is \(0\), which is less than 1, the series \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \) converges. Therefore, the series of absolute values converges.
05

Conclude Absolute Convergence

Since the series \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \) converges, our original series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \) converges absolutely.

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Key Concepts

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

Alternating Series
An alternating series is a series whose terms alternate in sign. In the series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \), the \((-100)^n\) factor causes each term to switch between positive and negative. This creates a back-and-forth effect in the progression of the series, which is characteristic of alternating series.

Alternating series can sometimes converge even when they do not converge absolutely. A key tool for testing the convergence of alternating series is the Alternating Series Test. This test states that an alternating series \( \sum (-1)^n a_n \) will converge if the following two conditions are met:
  • The terms \( a_n \) are decreasing in absolute value, meaning \( |a_{n+1}| \leq |a_n| \).
  • The limit of \( a_n \) as \( n \to \infty \) is zero: \( \lim_{n \to \infty} a_n = 0 \).
When these criteria are satisfied, the alternating series converges. However, in our analysis above, we are more broadly interested in absolute convergence.
Absolute Convergence
Absolute convergence refers to an infinite series \( \sum a_n \) that converges even when all its terms are replaced by their absolute values forming \( \sum |a_n| \). When a series converges absolutely, it means the corresponding non-alternating series \( \sum |a_n| \) also converges.

For example, in the series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \), to check absolute convergence, we consider \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \). If this series converges, then our original series converges absolutely. Knowing this is helpful because absolute convergence assures convergence of series by straightforward manipulation of terms, contrary to some series that only satisfy conditional convergence.
Ratio Test
The Ratio Test is a popular method for testing convergence of series, especially those involving factorials or exponentials. It is particularly effective for series in the form \( \sum a_n \) where \( a_n \) involves terms like factorials or powers. The test assesses the limit of the absolute ratio of successive terms:
\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

For the series \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \), the ratio becomes: \[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{100}{n+1} \]
Evaluating the limit gives:
\[ \lim_{n \to \infty} \frac{100}{n+1} = 0 \]
As the limit \( 0 \) is less than 1, the series \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \) converges according to the Ratio Test. Therefore, the original series including the alternating factor converges absolutely.
Factorial Series
Factorials feature prominently in series when one factor of each term is expressed as \( n! \). A factorial series like \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \) has terms that grow rapidly in size due to the factorial in the denominator \( n! \) quickly overpowering the constant raised to a power.

This typical quick growth of \( n! \) makes many such series converge. In our exercise, the presence of the factorial in \( n! \) ensures the terms of the unwanted portion of the ratio shrink dramatically as \( n \) increases, satisfying the Ratio Test.
  • **Factorial Growth:** In the denominator, \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \), grows faster than any polynomial or exponential in the numerator.
Due to this fast growth, factorials dominate the convergence properties of the series they are part of, often leading to absolute convergence in series like ours.

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Most popular questions from this chapter

Uniqueness of limits Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2}\)

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}.$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

Which of the series in Exercises 13 46 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.) $$ \sum_{n=1}^{\infty} \frac{n}{n^{2}+1} $$

Estimating Pi About how many terms of the Taylor series for \(\tan ^{-1} x\) would you have to use to evaluate each term on the right- hand side of the equation \begin{equation} \pi=48 \tan ^{-1} \frac{1}{18}+32 \tan ^{-1} \frac{1}{57}-20 \tan ^{-1} \frac{1}{239} \end{equation} with an error of magnitude less than \(10^{-6}\) ? In contrast, the convergence of \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)\) to \(\pi^{2} / 6\) is so slow that even 50 terms will not yield two-place accuracy.

\begin{equation} \begin{array}{l}{\text { a. Use Taylor's formula with } n=2 \text { to find the quadratic }} \\ {\text { approximation of } f(x)=(1+x)^{k} \text { at } x=0(k \text { a constant })} \\ {\text { b. If } k=3, \text { for approximately what values of } x \text { in the interval }} \\ {[0,1] \text { will the error in the quadratic approximation be less }} \\ {\text { than } 1 / 100 ?}\end{array} \end{equation}

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