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Which of the series in Exercises \(15 - 48\) converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } ( n ! ) ^ { 2 } } { ( 2 n ) ! } $$

Short Answer

Expert verified
The series converges absolutely by the ratio test.

Step by step solution

01

Analyze the Series

We are given the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (n!)^2}{(2n)!} \). This is an alternating series due to the term \((-1)^{n+1}\).
02

Absolute Convergence Check

To check for absolute convergence, we need to test if \( \sum_{n=1}^{\infty} \left| \frac{(n!)^2}{(2n)!} \right| \) converges. Since the absolute value of \((-1)^{n+1}\) is 1, the series we consider is \( \sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} \).
03

Use the Ratio Test for Absolute Convergence

We apply the ratio test to \( \sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} \). Consider the limit \[ \lim_{n \to \infty} \left| \frac{\left((n+1)!\right)^2}{(2n+2)!} \div \frac{(n!)^2}{(2n)!} \right|.\] Simplifying this gives \[ \lim_{n \to \infty} \frac{(n+1)^2}{(2n+1)(2n+2)}.\]
04

Evaluate the Limit for Ratio Test

Evaluating the limit, we get \( \lim_{n \to \infty} \frac{(n+1)^2}{(4n^2+6n+2)} = \lim_{n \to \infty} \frac{1}{4} = \frac{1}{4} \). Since \( \frac{1}{4} < 1 \), the series 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 one where the terms alternate in sign, often because of a factor like \((-1)^{n+1}\). The key feature of alternating series is that they can converge even if their absolute values diverge. This is particularly because the terms flip between positive and negative, allowing them to potentially balance out.When analyzing an alternating series, we often utilize the **Alternating Series Test**. This test states that a series \(\sum (-1)^n a_n\) converges if:
  • The absolute value of the terms \(|a_n|\) decreases steadily, i.e., \(a_{n+1} \leq a_n\) for all n.
  • The limit of \(|a_n|\) as n approaches infinity is zero, i.e., \(\lim_{{n \to \infty}} a_n = 0\).
In the context of absolute convergence, however, we look beyond these alternating signs and check if the series converges without the alternating factor.
Absolute Convergence
Absolute convergence means that the series \(\sum a_n\) converges even when all terms are made positive, forming \(|a_n|\). If a series converges absolutely, it means it converges completely regardless of the term signs.To determine absolute convergence, we often take the absolute value of each term. If the resulting series converges, the original series is said to converge absolutely.In the exercise provided, the series originally includes the factor \((-1)^{n+1}\), which makes it an alternating series. By removing this sign-changing factor, we analyze the series:\[\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}\]If this absolute series converges through any test (like the ratio test), it implies that our original alternating series converges absolutely.
Ratio Test
The Ratio Test is a powerful method used to determine the convergence of a series. It is particularly useful for sequences involving factorials or powers. With the Ratio Test, we evaluate:The limit:\[L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right|\]Where \(L\) will decide the behavior of the series:
  • If \(L < 1\), the series converges.
  • If \(L > 1\), the series diverges.
  • If \(L = 1\), the test is inconclusive.
In our exercise, the Ratio Test is applied to the absolute series \(\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}\). Calculating:\[\lim_{n \to \infty} \frac{(n+1)^2}{(4n^2+6n+2)} = \frac{1}{4}\]Since \( rac{1}{4} < 1\), the series converges absolutely.
Factorials in Series
Factorials often appear in series, providing unique challenges and opportunities for analysis. A factorial, expressed as \(n!\), is the product of all positive integers less than or equal to \(n\). Factorials grow extremely fast as \(n\) increases, which can greatly influence the convergence of a series.In the series under consideration, both the numerator and the denominator involve factorials:\[(n!)^2 \, \text{and} \, (2n)!\]The presence of \(n!\) in the numerator and \( (2n)!\) in the denominator often results in rapidly decreasing terms, especially for large \(n\), due to the fact that \( (2n)!\) grows significantly faster than \(n!\).Understanding the behavior of factorials in series helps in identifying appropriate tests for convergence, such as the Ratio Test, which works efficiently with these forms because of the way relative growth rates affect the series' terms.

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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 \(L_{1}=L_{2}\)

The (second) second derivative test Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}$$ to establish the following test. \begin{equation} \begin{array}{l}{\text { Let } f \text { have continuous first and second derivatives and suppose }} \\ {\text { that } f^{\prime}(a)=0 . \text { Then }} \\ {\text { a. } f \text { has a local maximum at } a \text { if } f^{\prime \prime} \leq 0 \text { throughout an interval }} \\ {\text { whose interior contains } a \text { ; }} \\ {\text { b. } f \text { has a local minimum at } a \text { if } f^{\prime \prime} \geq 0 \text { throughout an interval }} \\ {\quad \text { whose interior contains } a \text { . }}\end{array} \end{equation}

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.

Use a CAS to perform the following steps for the sequences. 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 \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\frac{8^{n}}{n !} $$

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.) $$ \sum_{n=1}^{\infty} \operatorname{sech} n $$

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