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Chapter 11: Problem 37

Which of the series \(\sum_{n=1}^{\infty} a_{n}\) defined by the formulas in Exercises \(27-38\) converge, and which diverge? Give reasons for your answers. $$ a_{n}=\frac{2^{n} n ! n !}{(2 n) !} $$

Short Answer

Expert verified
The series converges.

Step by step solution

01

Series Definition and Objective

We are given a series \( \sum_{n=1}^{\infty} a_{n} \) with \( a_n = \frac{2^n n! n!}{(2n)!} \). Our objective is to find whether this series converges or diverges.
02

Stirling's Approximation

To help simplify factorials, we use Stirling's approximation: \( n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \). This will help us understand the behavior of \( a_n \) as \( n \to \infty \).
03

Simplify Using Stirling's Approximation

Apply Stirling's approximation to each factorial in \( a_n \): \[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \] \[ (2n)! \approx \sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n} \] Plug these into \( a_n = \frac{2^n n! n!}{(2n)!} \).
04

Evaluate Asymptotic Behavior of Terms

Substitute our approximations: \[ a_n \approx \frac{2^n (\sqrt{2\pi n} (\frac{n}{e})^n)^2}{\sqrt{4\pi n} (\frac{2n}{e})^{2n}} \] Simplifying the multiplicative terms and fractions gives: \[ a_n \approx \frac{2^n \cdot 2\pi n \cdot (\frac{n}{e})^{2n}}{\sqrt{4\pi n} (\frac{2^{2n} n^{2n}}{e^{2n}} )} \]Further simplify: \[ a_n \approx \frac{2^{n-1.5} \pi n}{2\pi \cdot 2^{2n} } \propto \frac{1}{n^{3/2}} \] The dominant term \( \frac{1}{n^{3/2}} \) suggests the series converges.
05

Convergence Test

Apply the convergence test for p-series which states the series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges for \( p > 1 \). Here, we have \( p = 3/2 \), thus the series \( \sum_{n=1}^{\infty} a_n \) converges.

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Key Concepts

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

Stirling's Approximation
Stirling's Approximation is a mathematical formula used to approximate factorials. For large numbers, calculating exact factorial values becomes difficult. Stirling鈥檚 Approximation provides a practical way to make these calculations more manageable.The approximation is given by:\[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \]This formula shows that the factorial of a number \( n \) can be expressed as a combination of exponential growth \( (\frac{n}{e})^n \) and a superimposed square root and constant factor \( \sqrt{2\pi n} \). In the problem, this approximation is used to simplify the expression \( a_n = \frac{2^n n! n!}{(2n)!} \), to explore whether the series converges. By replacing the factorial terms with Stirling鈥檚 form, manageable expressions for large \( n \) can be computed. This simplifies analytical work, allowing us to understand behavior as \( n \to \infty \).
P-Series Test
Analyzing whether an infinite series converges can be challenging. The P-Series Test is a straightforward method for certain series types. A p-series has the form:\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]where \( p \) is a positive constant. The test states:- The series converges if \( p > 1 \).- The series diverges if \( p \leq 1 \).In the solution, we found that our transformed series resembles a p-series with \( p = \frac{3}{2} \). Since \( \frac{3}{2} > 1 \), the P-Series Test confirms the series converges. This simple yet effective principle helps streamline convergence assessments for series analysis in calculus.
Factorial Simplification
Factorial simplification is essential for analyzing series like the given one, especially when using Stirling's Approximation. This is particularly relevant when dealing with complex series expressions that include high factorial terms.Factorials grow very quickly, making it impractical to handle them exactly for large numbers. In the step-by-step solution, Stirling's Approximation allowed the simplification of:\[ n! \text{ and } (2n)! \]These simplifications transformed the difficult-to-use original form into one where terms cancel or reduce easily.For example, by substituting the approximated forms: \[ \frac{2^n n! n!}{(2n)!} \approx \frac{2^n (\sqrt{2\pi n} (\frac{n}{e})^n)^2}{\sqrt{4\pi n} (\frac{2n}{e})^{2n}} \]We end up with more generic expressions, revealing more about the series' behavior. This leads us to a manageable power form \( \frac{1}{n^{3/2}} \), which we can readily analyze with the P-Series Test. Overall, factorial simplifications help unlock otherwise compounded arithmetic and algebraic problems.

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Most popular questions from this chapter

The Cauchy condensation test says: Let \(\left\\{a_{n}\right\\}\) be a nonincreasing sequence \(\left(a_{n} \geq a_{n+1} \text { for all } n\right)\) of positive terms that converges to \(0 .\) Then \(\sum a_{n}\) converges if and only if \(\sum 2^{n} a_{2 n}\) converges. For example, \(\sum(1 / n)\) diverges because \(\Sigma 2^{n} \cdot\left(1 / 2^{n}\right)=\sum 1\) diverges. Show why the test works.

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) 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 \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{1}=1, \quad a_{n+1}=a_{n}+(-2)^{n} $$

Outline of the proof of the Rearrangement Theorem (Theo- rem 17\()\) a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n},\) and let \(s_{k}=\sum_{n=1}^{k} a_{n}\) . Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\) $$\sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \quad\( and \)\quad\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2}$$ since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\},\) there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3},\) then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1} .\) Therefore, if \(n \geq N_{3}\) $$\begin{aligned}\left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned}$$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\)

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x) .\) Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\) , $$ \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right) $$
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