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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblem 10.7.39
Chapter 10, Problem 10.7.39

In Exercises 37–42, find the series’ radius of convergence.
∑ (from n = 1 to ∞) [ (n!)² / (2ⁿ (2n)!) ] xⁿ

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Identify the general term of the series as \( a_n = \frac{(n!)^2}{2^n (2n)!} x^n \). Our goal is to find the radius of convergence \( R \) of the power series \( \sum_{n=1}^\infty a_n \).
Use the Ratio Test to find the radius of convergence. Consider the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). The radius of convergence \( R \) is given by \( R = \frac{1}{L} \) if the limit exists.
Write out the ratio \( \left| \frac{a_{n+1}}{a_n} \right| = \frac{\frac{((n+1)!)^2}{2^{n+1} (2(n+1))!} |x|^{n+1}}{\frac{(n!)^2}{2^n (2n)!} |x|^n} = \frac{((n+1)!)^2}{(n!)^2} \cdot \frac{2^n}{2^{n+1}} \cdot \frac{(2n)!}{(2(n+1))!} \cdot |x| \).
Simplify the factorial expressions: \( \frac{((n+1)!)^2}{(n!)^2} = (n+1)^2 \), \( \frac{2^n}{2^{n+1}} = \frac{1}{2} \), and \( \frac{(2n)!}{(2n+2)!} = \frac{1}{(2n+1)(2n+2)} \). Substitute these back into the ratio.
Express the limit \( L = \lim_{n \to \infty} (n+1)^2 \cdot \frac{1}{2} \cdot \frac{1}{(2n+1)(2n+2)} \cdot |x| \). Analyze this limit to find \( L \) in terms of \( |x| \), then solve \( L < 1 \) to determine the radius of convergence \( R \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Radius of Convergence

The radius of convergence of a power series is the distance from the center of the series within which the series converges absolutely. It can be found using tests like the Ratio Test or Root Test, and it determines the interval on the x-axis where the series represents a valid function.
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Radius of Convergence

Ratio Test for Series Convergence

The Ratio Test involves taking the limit of the absolute value of the ratio of consecutive terms in a series. If this limit is less than one, the series converges; if greater than one, it diverges. This test is especially useful for power series to find the radius of convergence.
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Ratio Test

Factorials and Their Growth

Factorials (n!) grow very rapidly and often dominate the behavior of series terms. Understanding how factorial expressions behave, especially in ratios like (n!)²/(2n)!, is crucial for simplifying terms and applying convergence tests effectively.
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Factorials
Related Practice
Textbook Question

In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.

∑ (from n = 2 to ∞) [(ln n / n)³]

Textbook Question

Determining Convergence or Divergence

Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.

∑ (from n=1 to ∞) sin (1/n)

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Textbook Question

Recursively Defined Sequences

In Exercises 101–108, assume that each sequence converges and find its limit.

a₁ = 5,aₙ₊₁ = √(5aₙ)

Textbook Question

Finding nth Partial Sums

In Exercises 1–6, find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges.

2 + (2/3) + (2/9) + (2/27) + … + (2 / 3ⁿ⁻¹) + …

Textbook Question

Finding a Sequence’s Formula

In Exercises 13–30, find a formula for the nth term of the sequence.

2, 6, 10, 14, 18, …Every other even positive integer

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Textbook Question

Using the Root Test

In Exercises 9–16, use the Root Test to determine if each series converges absolutely or diverges.

∑(from n=1 to ∞) [4ⁿ / (3n)ⁿ]