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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 8 - Problem 16re
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 8 - Problem 16re

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# Answer: Evaluating series Evaluate the following infinite

ISBN: 9780321570567 2

## Solution for problem 16RE Chapter 8

Calculus: Early Transcendentals | 1st Edition

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Problem 16RE

12-20. Evaluating series Evaluate the following infinite series or state that the series diverges.

$$\sum_{k=2}^{\infty}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k-1}}\right)$$

Step-by-Step Solution:
Step 1 of 3

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 S n 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 + k k ) and ( U - Vk k ) areconvergent . If C = / 0 , then U k and V k both converge or both diverge . Step-2a + ar + ar +ar +..................... is an infinite geometric progression.Here 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……………(1) r=0 NOTE ; The infinite geometric series x converges if |x| < 1 , and diverges if |x| 1 r =0\n Step-3 1 1 The given sequence is S k ( k k 1) k =2 n Then S n ( 1 1 ) k =2 k k 1 1 1 1 1 1 = ( 2 -1 + 3 - 2 +..... …+ n - n1 ) = ( 1 - 1) n Therefore , S =n 1 - 1 n 1Now , we have to find the value of lim S = limn( n - 1) n n 1 = - 1 , since as n , then n. Therefore , ( 1 1 ) = -1 . k =2 k k 1 From the step(1) , step(2) , it is clear that the given infinite series is convergent . Becausethe value approaches to ‘-1’ (finite value). Therefore , ( 1 1 ) is convergent series. k =2 k k 1

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