Calculus 3 Task 4.docx - Trishann Russell WGU Student Number 000779161 Calculus III \u2013 C656 Task 4 Alternating Series Converge if an >0 a1 >a 2> a3 \u2026

# Calculus 3 Task 4.docx - Trishann Russell WGU Student...

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Trishann Russell WGU Student Number 000779161 Calculus III – C656 Task 4 Alternating Series Converge if: 1. a n > 0 2. a 1 > a 2 > a 3 . . 3. lim n→∞ a n = 0 j = 1 ( 1 ) j 1 j + c = j = 1 ( 1 ) j 1 j + 29 where a j = ¿ f(j) = 1 ( j + 29 ) f(j) is positive if both numerator and denominator have the same sign. In 1 j + 29 the numerator is a positive constant. Therefore the denominator needs to be positive. (√(j+29)) =0 ( j + 29 ) ¿¿ 2 = 0 2 ( j + 29 ) = 0 j =− 29 The denominator will be positive for all j ¿ -29 since we are only looking at [1, ∞) the denominator will always be greater than or equal to √(1+29)=√30 so both numerator and denominator will always be positive therefore f(j) will always be positive. If it is decreasing f’(j) must be negative between [1, ∞) f ( j ) = 1 j + 29 = 1 ( j + 29 ) 1 2 = ( j + 29 ) 1