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Which of the series in Exercises \(11-44\) 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.

Step by step solution

01

Identify the series type

The series given is \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n!)^{2}}{(2n)!} \). This is an alternating series, as indicated by the presence of \((-1)^{n+1}\). In an alternating series, we analyze absolute convergence and simple convergence separately.
02

Test for Absolute Convergence

To check for absolute convergence, consider the series of absolute values: \( \sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2n)!} \). We apply the ratio test:
03

Step 2a: Apply Ratio Test

The ratio test is used to determine the nature of the series. Compute:\[L = \lim_{n \to \infty} \left| \frac{(n+1)!^{2}}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^{2}} \right|\] Simplifying this ratio:\[\L = \lim_{n \to \infty} \left| \frac{(n+1)^{2}(n!)^{2}}{(2n+2)(2n+1)(2n)!} \cdot \frac{(2n)!}{(n!)^{2}} \right|\] Cancelling terms yields:\[\L = \lim_{n \to \infty} \left| \frac{(n+1)^{2}}{(2n+2)(2n+1)} \right|\]
04

Step 2b: Evaluate the Limit

Evaluate the limit:\[L = \lim_{n \to \infty} \frac{(n+1)^{2}}{4n^{2} + 6n + 2}\]Divide the numerator and the denominator by \(n^{2}\):\[L = \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{4 + \frac{6}{n} + \frac{2}{n^2}} = \frac{1}{4}\] Since \(L < 1\), the series of absolute values converges.
05

Confirm Convergence of Original Series

To check the convergence of the original alternating series, apply the Alternating Series Test. Consider the non-alternating part:\( a_n = \frac{(n!)^{2}}{(2n)!} \). The series converges if:1. \(a_n\) is decreasing;2. \(\lim_{n \to \infty} a_n = 0\).
06

Step 3a: Analyze Decreasing Nature

To show \(a_n\) is decreasing, note the simplicity of the ratio used in Step 2a. The expression \(\frac{(n+1)^{2}}{(2n+2)(2n+1)}\) indicates that \(a_n\) is decreasing because the ratio is less than 1.
07

Step 3b: Limit of Terms

Compute the limit:\[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{(n!)^{2}}{(2n)!} = 0\]By the Alternating Series Test, the original series converges.

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Key Concepts

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

Absolute Convergence
When we talk about absolute convergence, we mean that a series converges when you take the absolute value of each of its terms. This is a stronger form of convergence because if a series converges absolutely, it will also converge in the usual sense, but not necessarily the other way around.

To check for absolute convergence, you consider the series of absolute values. In the given exercise, the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n!)^{2}}{(2n)!} \) is transformed to \( \sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2n)!} \).

After effectively using the Ratio Test, we find that the limit \( L \) comes out to be \( \frac{1}{4} \), which is less than 1. This indicates the series of absolute values converges, thus confirming that the original series converges absolutely.
Ratio Test
The Ratio Test is a powerful tool used to determine if an infinite series converges or diverges. It involves taking the limit of the ratio of successive terms and is particularly useful for series whose terms contain factorials or exponential functions.

For the series \( \sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2n)!} \), we calculate the limit\[L = \lim_{n \to \infty} \left| \frac{(n+1)!^{2}}{(2n+2)!} \times \frac{(2n)!}{(n!)^{2}} \right|\]By simplifying this expression, the limit \( L = \frac{1}{4} \) is found. Since this limit is less than 1, the Ratio Test tells us that the series converges.

The key advantage of the Ratio Test is in situations involving factorials as they simplify nicely, leading to easy-to-evaluate limits.
Alternating Series Test
The Alternating Series Test is used to determine the convergence of series where terms alternate in sign. For a series to pass this test:
  • The absolute value of the terms should be decreasing.
  • The limit of the terms as \( n \to \infty \) must be 0.
In our original series, \( a_n = \frac{(n!)^2}{(2n)!} \), we verify that:
  • Each term \( a_n \) is smaller than the previous one, thanks to the expression \( \frac{(n+1)^2}{(2n+2)(2n+1)} \) being less than 1.
  • The limit of \( a_n \) as \( n \to \infty \) is indeed 0.
By fulfilling these conditions, the Alternating Series Test confirms that the series converges. This method is very handy for alternating series, providing a simple yet effective way to ensure convergence when absolute convergence is not evident.

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Most popular questions from this chapter

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+y=0, \quad y^{\prime}(0)=0 \text { and } y(0)=1 $$

What is wrong here?: Multiply both sides of the alternating harmonic series $$S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+$$ $$\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\cdots$$ by 2 to get $$2 S=2-1+$$ $$\frac{2}{3}-\frac{1}{2}+\frac{2}{5}-\frac{1}{3}+\frac{2}{7}-\frac{1}{4}+\frac{2}{9}-\frac{1}{5}+\frac{2}{11}-\frac{1}{6}+\cdots$$ Collect terms with the same denominator, as the arrows indicate, to arrive at $$2 S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$$ The series on the right-hand side of this equation is the series we started with. Therefore, \(2 S=S\), and dividing by \(S\) gives \(2=1\). (Source: "Riemann's Rearrangement Theorem" by Stewart Galanor, Mathematics Teacher, Vol. 80, No. 8,1987, pp. \(675-681 .)\)

Two complex numbers \(a+i b\) and \(c+i d\) are equal if and only if \(a=c\) and \(b=d .\) Use this fact to evaluate $$ \int e^{a x} \cos b x d x \text { and } \int e^{a x} \sin b x d x $$ from $$ \int e^{(a+i b) x} d x=\frac{a-i b}{a^{2}+b^{2}} e^{(a+i b) x}+C $$ where \(C=C_{1}+i C_{2}\) is a complex constant of integration.

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime}-y=x, \quad y(0)=0 $$

Compound interest, deposits, and withdrawals If you invest an amount of money \(A_{0}\) at a fixed annual interest rate \(r\) compounded \(m\) times per year, and if the constant amount \(b\) is added to the account at the end of each compounding period (or taken from the account if \(b<0 ),\) then the amount you have after \(n+1\) compounding periods is $$ A_{n+1}=\left(1+\frac{r}{m}\right) A_{n}+b $$ a. If \(A_{0}=1000, r=0.02015, m=12,\) and \(b=50\) , calculate and plot the first 100 points \(\left(n, A_{n}\right) .\) How much money is in your account at the end of 5 years? Does \(\left\\{A_{n}\right\\}\) converge? Is \(\left\\{A_{n}\right\\}\) bounded? b. Repeat part (a) with \(A_{0}=5000, r=0.0589, m=12,\) and \(b=-50 .\) c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5\(\%\) annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have \(20,000\) dollars? What if the CD earns 6.25\(\% ?\) d. It can be shown that for any \(k \geq 0\) , the sequence defined recursively by Equation \((1)\) satisfies the relation $$ A_{k}=\left(1+\frac{r}{m}\right)^{k}\left(A_{0}+\frac{m b}{r}\right)-\frac{m b}{r} $$ For the values of the constants \(A_{0}, r, m,\) and \(b\) given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation \((2)\) satisfy the recursion formula in Equation ( 1\()\) .

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