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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 8 - Problem 12re
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 8 - Problem 12re

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# Evaluating series Evaluate the following infinite series

ISBN: 9780321570567 2

## Solution for problem 12RE Chapter 8

Calculus: Early Transcendentals | 1st Edition

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Problem 12RE

Evaluating series Evaluate the following infinite series or state that the series diverges.

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STEP_BY_STEP SOLUTION Step-1 Definition ; A series is said to be convergent if it approaches some limit .Formally , the infinite series a n is convergent if the sequence of partial sums n=1 n Sn= a is convergent . Conversely , a series is divergent if the k=1 ksequence ofpartial sums is divergent.NOTE : The terms grow without bound , so the sequence does not converge. If U k and V k are convergent series , then ( U + Vk k ) and ( U - Vk k ) areconvergent . If C = / 0 , then U k and V k both converge or both diverge . Step-2 9 kNow , the given infinite series is an= ( ) 10 k = 1 9 k 9 9 2 9 3 9 4 ( )10 = 10 + ( 10+( ) +10) + ……10……. k = 1 9 Here , the given series in geometric progression , and the common ratio is 10 < 1 . In the above series first term a = 9 , and 10 The common ratio ‘r’ = 10 ……………..(1) NOTE ; a + ar + ar +ar +..................... is an infinite geometric progression.\nHere the first term is ‘a’ , and the common ratio is ‘r’ n a If |r| < 1 , then the sum of the series ar = 1 r……….(2) r=0 9 k 9 9 k ( )10 = ( 10) ( 10 k = 1 k = 0 Using , the above note ; the sum of the infinite series is 9 9 k 10 ( 10) ( 10= 1 9 , since from (1) , (2) . k = 0 10 10 10 = 10 9= 1 10 10 = 9 ( 10 ) = 9. 10 1 9 k Therefore , ( )10 = 9 . k = 1

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