Explanation of Solution Given: 1.234567 = 1.234567567567... Result used: The geometric series >) `ar"-1 (or) a + ar + ar² + is convergent if n=1 |r| < 1 and its sum is -"-, where a is the first term and ris the common ratio of the series. Calculation: Rewrite the number and express 1.234567 as follows, 1.234567 = 1.234 + 0.000567 1.234 + (0.00 67+ 0.000000567+ ·· ·) = 1.234 + 106 567 567 +...) 10° 567 (1) 1.234567 = 1.234 + 103n n=2 567 is geometric series with first term of the series is a Here, 567 103n n=: 106 and common ratio is r = 103 Since |r| < 1 and the Result stated above, the geometric series 567 is 103n n=2 convergent.
Explanation of Solution Given: 1.234567 = 1.234567567567... Result used: The geometric series >) `ar"-1 (or) a + ar + ar² + is convergent if n=1 |r| < 1 and its sum is -"-, where a is the first term and ris the common ratio of the series. Calculation: Rewrite the number and express 1.234567 as follows, 1.234567 = 1.234 + 0.000567 1.234 + (0.00 67+ 0.000000567+ ·· ·) = 1.234 + 106 567 567 +...) 10° 567 (1) 1.234567 = 1.234 + 103n n=2 567 is geometric series with first term of the series is a Here, 567 103n n=: 106 and common ratio is r = 103 Since |r| < 1 and the Result stated above, the geometric series 567 is 103n n=2 convergent.
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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In the image attached, the geometric series I need to solve begins with "2" (n = 2). I'm not sure why it begins at 2, and not 1. Do you know why this series begins with 2? Let me know if you need more clarification!
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