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Chapter 1: Problem 21

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{5^{n}(n !)^{2}}{(2 n) !}$$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

- Write down the ratio test formula

To use the ratio test, consider the series \ \( \sum_{n=0}^{\infty} a_n \). The ratio test involves computing \ \(\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \). If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive.
02

- Express \(a_n\) for the given series

For the given series, identify \ \(a_n = \frac{5^n (n!)^2}{(2n)!} \).
03

- Find \(a_{n+1}\)

Substitute \ \(n+1\) in place of \ \(n\) in the expression for \ \(a_n\), obtaining \ \(a_{n+1} = \frac{5^{n+1} ((n+1)!)^2}{(2(n+1))!} = \frac{5^{n+1} ((n+1)!)^2}{(2n+2)!} \).
04

- Form the ratio \(\frac{a_{n+1}}{a_n}\)

Compute the ratio: \ \[\frac{a_{n+1}}{a_n} = \frac{5^{n+1} ((n+1)!)^2}{(2n+2)!} \cdot \frac{(2n)!}{5^n (n!)^2} = 5 \cdot \frac{(n+1)^2}{(2n+2)(2n+1)}.\] Simplify the ratio to: \ \[\frac{a_{n+1}}{a_n} = 5 \cdot \frac{n^2 + 2n + 1}{4n^2 + 6n + 2}.\]
05

- Take the limit as \(n\) approaches infinity

Evaluate the limit: \ \[\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} 5 \cdot \frac{n^2 + 2n + 1}{4n^2 + 6n + 2} = 5 \cdot \frac{1}{4} = \frac{5}{4}.\]
06

- Conclude the test

Since the limit \(\frac{5}{4} > 1\), by the ratio test, the given series \(\sum_{n=0}^{\infty} \frac{5^n (n!)^2}{(2n)!}\) diverges.

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Key Concepts

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

infinite series
An infinite series is the sum of the terms of an infinite sequence. Imagine you have a sequence of numbers: a1, a2, a3, and so on, continuing without end. When you add these numbers together, you get an infinite series: a1 + a2 + a3 +.... These series can behave in different ways. Sometimes, the sum reaches a finite number even though there are infinitely many terms. Other times, the series grows without bound.
convergence and divergence
In the context of infinite series, convergence and divergence are crucial concepts. When we say a series converges, we mean the sum of the series approaches a specific, finite number as more terms are added. Conversely, if a series diverges, the sum either goes to infinity, negative infinity, or does not approach any specific number as terms are added. The behavior of a series (converging or diverging) helps us understand if the sum results in a meaningful number or not.
limit comparison test
The limit comparison test is a method used to determine whether an infinite series converges or diverges. It is particularly useful when dealing with series that are tough to handle directly. This test compares the given series to a known benchmark series with well-understood convergence or divergence behavior. Given two positive term series \textteam.tex{\frac{\textlambda \(\frac{a_n}{b_n}}\)}, where the given series (`a_n`) is compared against a known series (`b_n`). If the ratio of \frac{\textlambda \(\frac{a_n}{b_n}}\). If this ratio is a positive finite number, both series either converge or both diverge. This technique can significantly simplify the process of determining a series' behavior.

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Most popular questions from this chapter

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\frac{x}{\sin x}$$

(a) It is clear that you (or your computer) can't find the sum of an infinite series just by adding up the terms one by one. For example, to get \(\zeta(1.1)=\) \(\sum_{n=1}^{\infty} 1 / n^{1.1}(\text { see Problem } 15.22)\) with error \(<0.005\) takes about \(10^{33}\) terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed \(N\) terms, the sum \(R_{N}\) of the rest of the series is between \(I_{N}=\int_{N}^{\infty} a_{n} d n\) and \(I_{N+1}=\int_{N+1}^{\infty} a_{n} d n\) (b) Find the integrals in (a) for the \(\zeta(1.1)\) series and verify the claimed number of terms needed for error \(<0.005 .\) Hint: Find \(N\) such that \(I_{N}=0.005 .\) Also find upper and lower bounds for \(\zeta(1.1)\) by computing \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N}^{\infty} n^{-1.1} d n\) and \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N+1}^{\infty} n^{-1.1} d n\) where \(N\) is far less than \(10^{33} .\) Hint: You want the difference between the upper and lower limits to be about \(0.005 ;\) find \(\mathrm{N}\) so that term \(a_{N}=0.005\).

(a) Using computer or tables (or see Chapter 7 , Section 11 ), verify that \(\sum_{n=1}^{\infty} 1 / n^{2}=\) \(\pi^{2} / 6=1.6449+,\) and also verify that the error in approximating the sum of the series by the first five terms is approximately 0.1813 . (b) \(\quad\) By computer or tables verify that \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)(1 / 2)^{n}=\pi^{2} / 12-(1 / 2)(\ln 2)^{2}=\) \(0.5822+,\) and that the sum of the first five terms is \(0.5815+.\) (c) Prove theorem (14.4). Hint: The error is \(\left|\sum_{N+1}^{\infty} a_{n} x^{n}\right| .\) Use the fact that the absolute value of a sum is less than or equal to the sum of the absolute values. Then use the fact that \(\left|a_{n+1}\right| \leq\left|a_{n}\right|\) to replace all \(a_{n}\) by \(a_{N+1},\) and write the appropriate inequality. Sum the geometric series to get the result.

Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals (see Problem 16 ). $$\sum_{n=1}^{\infty} \frac{n}{n^{2}+4}$$

Find the Maclaurin series for the following functions. $$\arctan x=\int_{0}^{x} \frac{d u}{1+u^{2}}$$
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