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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 10.7.40
Chapter 10, Problem 10.7.40

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

Verified step by step guidance
1
Identify the general term of the series as \(a_n = \frac{n! x^n}{n^n}\).
Recall that the radius of convergence \(R\) can be found using the Root Test or the Ratio Test. Here, the Ratio Test is convenient, which involves evaluating \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
Write the ratio \(\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)! x^{n+1} / (n+1)^{n+1}}{n! x^n / n^n} \right| = |x| \cdot \frac{(n+1)!}{n!} \cdot \frac{n^n}{(n+1)^{n+1}}\).
Simplify the factorial and powers: \(\frac{(n+1)!}{n!} = n+1\), so the ratio becomes \(|x| (n+1) \frac{n^n}{(n+1)^{n+1}}\).
Rewrite \(\frac{n^n}{(n+1)^{n+1}}\) as \(\frac{n^n}{(n+1)^n (n+1)} = \frac{1}{n+1} \left( \frac{n}{n+1} \right)^n\), then simplify the entire expression and take the limit as \(n \to \infty\) to find 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 determines the interval on the x-axis where the series sums to a finite value. Finding this radius helps understand the domain of validity for the series representation.
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Radius of Convergence

Ratio Test for Series Convergence

The ratio test is a method to determine the convergence of an infinite series by examining the limit of the absolute value of the ratio of consecutive terms. For power series, it is commonly used to find the radius of convergence by setting the limit of |a_{n+1}/a_n| equal to 1/|x|.
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Ratio Test

Stirling's Approximation

Stirling's approximation provides an estimate for factorials, especially useful for large n, stating n! ≈ √(2πn) (n/e)^n. This approximation simplifies factorial expressions in series terms, making it easier to analyze limits and apply convergence tests.
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Determining Error and Relative Error
Related Practice
Textbook Question

Determining Convergence or Divergence

In Exercises 17–46, use any method to determine whether the series converges or diverges. Give reasons for your answer.

∑(from n=1 to ∞) [(-1)ⁿ n² e⁻ⁿ]

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 ∞) (1 − n) / n2ⁿ

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Textbook Question

Intervals of Convergence

In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?

∑ (from n = 1 to ∞) [ (x − 1)ⁿ / (n³ 3ⁿ) ]

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Textbook Question

Use series to evaluate the limits in Exercises 29–40.

37. lim (x → 0) ln(1 + x²) / (1 - cos(x))

Textbook Question

Applying the Integral Test

Use the Integral Test to determine if the series in Exercises 1–12 converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied.

∑ (from n = 1 to ∞) 1 / n⁰·²

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Textbook Question

In Exercises 67–72, use the results of Exercises 63 and 64 to determine if each series converges or diverges.

∑(from n=2 to ∞) [(ln n)¹⁰⁰⁰ / n¹.⁰⁰¹]

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