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Find the general term of the series and use the ratio test to show that the series converges. 1.22 1.2.33 1.2.3.44 11+ + 13 135 1 3 5 7 O O O ● Uk= Uk= Uk = Uk = uk = (11k)! 1.3.5.. 5. (11k)! (2k-1) ' 1.3.5. .. (2k-1)=2 11k! 1 · 3 · 5 · · · ( 2 k − 1); P = 2 11k! 1 · 3 · 5 · · · ( 2 k − 1)! ' P 11k! 1.3.5 = 2 (2 k − 1) ³ º 1 11 72711 + +.

Find the general term of the series and use the ratio test to show that the series converges. 1.22 1.2.33 1.2.3.44 11+ + 13 135 1 3 5 7 O O O ● Uk= Uk= Uk = Uk = uk = (11k)! 1.3.5.. 5. (11k)! (2k-1) ' 1.3.5. .. (2k-1)=2 11k! 1 · 3 · 5 · · · ( 2 k − 1); P = 2 11k! 1 · 3 · 5 · · · ( 2 k − 1)! ' P 11k! 1.3.5 = 2 (2 k − 1) ³ º 1 11 72711 + +.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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View Policies Current Attempt in Progress Find the general term of the series and use the ratio test to show that the series converges. 122 1233 11+ + 13 135 O O O O O uk = Uk = uk = uk = uk = (11k)! 1.3.5. (2k-1)' (11k)! 1·3·5· · · (2k-1); P = 1/2 .. 11k! 1.3.5. 1.3.5.. 11k! Save for Later 1.3.5.. . 11k! . eTextbook and Media = 2 (2k-1) P = 2 1 2 (2k-1)! P (2k-1)= 12 + 1.2.3.44 1 3 5 7 . + Attempts: 0
Transcribed Image Text:View Policies Current Attempt in Progress Find the general term of the series and use the ratio test to show that the series converges. 122 1233 11+ + 13 135 O O O O O uk = Uk = uk = uk = uk = (11k)! 1.3.5. (2k-1)' (11k)! 1·3·5· · · (2k-1); P = 1/2 .. 11k! 1.3.5. 1.3.5.. 11k! Save for Later 1.3.5.. . 11k! . eTextbook and Media = 2 (2k-1) P = 2 1 2 (2k-1)! P (2k-1)= 12 + 1.2.3.44 1 3 5 7 . + Attempts: 0
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