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EXAMPLE 10 Discuss the convergence n! SOLUTION We observe that n (n-1) 3.2.1. ... an ² of the sequence Then, since 1/n → 0 theorem) ano. n. (n-1). n n. (n-1) n. n . . as 2 ... an n - 3.2.1 • n • n •n 3 an since the numerator is less than or equal to the denominator for the quantity in parenthesis. Therefore • n (1 -). ÷ n n!/n", where ² n we have (by the squeeze )

EXAMPLE 10 Discuss the convergence n! SOLUTION We observe that n (n-1) 3.2.1. ... an ² of the sequence Then, since 1/n → 0 theorem) ano. n. (n-1). n n. (n-1) n. n . . as 2 ... an n - 3.2.1 • n • n •n 3 an since the numerator is less than or equal to the denominator for the quantity in parenthesis. Therefore • n (1 -). ÷ n n!/n", where ² n we have (by the squeeze )

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter9: Sequences, Probability And Counting Theory
Section9.1: Sequences And Their Notations
Problem 69SE: Find a recursive formula for the sequence 1,0,1,1,0,1,1,0,1,1,0,1,1,... (Hint: find a pattern for an...
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EXAMPLE 10 Discuss the n! SOLUTION We observe that convergence n (n-1) - 3.2.1. an = of the sequence n. (n-1). Then, since 1/n → 0 theorem) ano. n. (n-1). n ² n O as ♦ 2 . an since the numerator is less than or equal to the denominator for the quantity in parenthesis. Therefore 3.2.1 n.n.n an 3.2 -) = /1 n • n • n n→D, n!/n", where ² n we have (by the squeeze
Transcribed Image Text:EXAMPLE 10 Discuss the n! SOLUTION We observe that convergence n (n-1) - 3.2.1. an = of the sequence n. (n-1). Then, since 1/n → 0 theorem) ano. n. (n-1). n ² n O as ♦ 2 . an since the numerator is less than or equal to the denominator for the quantity in parenthesis. Therefore 3.2.1 n.n.n an 3.2 -) = /1 n • n • n n→D, n!/n", where ² n we have (by the squeeze
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