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Chapter 11: Problem 44

Which of the series in Exercises \(39-44\) converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)} $$

Short Answer

Expert verified
The series converges by the ratio test.

Step by step solution

01

Recognize the Series Format

The given series is \( \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)} \). This involves a factorial-like expression in the numerator and denominator.
02

Simplify the Fraction

Each term of the series simplifies further since \( 1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1) = (2n-1)!! \) and \( 2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n) = (2n)!! \). The odd and even factorials relate as \( \frac{(2n-1)!!}{(2n)!!} = \frac{( (2n)! / 2^n n!)}{(2^n n!)} \). Hence, this fraction becomes \( \frac{1}{2^n} \).
03

Series Expression

The expression for each term simplifies to \( \frac{1}{2^n(3^n+1)} \). The general term is thus \( \frac{1}{6^n + 2^n} \).
04

Apply Ratio Test

For ratio test, compute \( \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{6^{n+1} + 2^{n+1}} \times \frac{6^n + 2^n}{1} \). As \( n \to \infty \), the limit of this ratio approaches \( \frac{1}{6} \).
05

Convergence Result by Ratio Test

Since \( \frac{1}{6} < 1 \), the series converges according to the ratio test.

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Key Concepts

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

Ratio Test
The ratio test is a powerful tool used to determine the convergence or divergence of a series. It works well, especially with series involving factorials, powers, or any form of exponential growth. To apply the ratio test, we consider the absolute value of the ratio between successive terms of a series. For a given series with terms \(a_n\), the ratio test involves calculating:\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]If the limit \(L\) exists, the series converges if \(L < 1\) and diverges if \(L > 1\). If \(L = 1\), the test is inconclusive, and other methods must be used. In our problem, this method shows that the series converges, which means its terms shrink fast enough for the infinite sum to have a finite value.
Infinite Series
An infinite series is the sum of an infinite sequence of numbers. It may look like this: \( \sum_{n=1}^{\infty} a_n \). Each term \( a_n \) is part of a sequence, and the series adds these terms together to possibly reach a sum. However, not all infinite series converge. A series is said to converge when the sum approaches a finite number as we add more terms.- If the partial sums (the sum of the first few terms) stay bounded and get closer to a specific value, the series converges.- If the partial sums do not approach a particular value, the series diverges.In this exercise, the use of the ratio test determined that our series converges, meaning it has a finite sum, despite its infinite nature.
Factorial Notation
Factorial notation is a way to express a product of consecutive integers. For example, \(n!\) (read as 'n factorial') represents the product \(n \times (n-1) \times (n-2) \times ... \times 1\). Factorials grow very fast with increasing \(n\), which makes them important in many calculations, such as probabilities and series solutions.In the series examined in the exercise, a double factorial notation \((2n-1)!!\) and \((2n)!!\) is used:- \((2n-1)!!\) indicates the product of all odd numbers up to \((2n-1)\).- \((2n)!!\) indicates the product of all even numbers up to \(2n\).The proper handling of factorial notation in series can simplify calculations significantly and is crucial in higher mathematics, such as calculus and combinatorics.

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Most popular questions from this chapter

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{x \rightarrow \infty}(x+1) \sin \frac{1}{x+1} $$

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} .

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 $$ \pi=48 \tan ^{-1} \frac{1}{18}+32 \tan ^{-1} \frac{1}{57}-20 \tan ^{-1} \frac{1}{239} $$ 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.

Which of the series 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鈥� convergence or divergence.) $$ \sum_{n=1}^{\infty} \operatorname{sech} n $$

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 \(L ?\) $$ a_{n}=123456^{1 / n} $$
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