Use the Alternating Series Test to determine whether the series converges sigma^
ID: 2865028 • Letter: U
Question
Use the Alternating Series Test to determine whether the series converges sigma^infinity_k=1 (-1)^k+1 k/6k^6+1 Let a_k represent the magnitude of the terms of the given series and f(x) be the function that generates a_k. Determine whether the terms of the series are no increasing in magnitude. Select the correct choice below and fill in the answer box in your choice. The terms of the series are increasing in magnitude. Since f (x) = > 0 for x > 1/Squareroot30,. the terms a_k increase for k greaterthanorequalto 1. 1/Squareroot30 the terms a_k decrease for k greaterthanorequalto:1. Choose the correct answer below. The series diverges The series convergesExplanation / Answer
f(x)=x/(6x6+1)
f '(x)=[1(6x6+1) -x(6*6x5+0)]/(6x6+1)2
f '(x)=[6x6+1 -36x6]/(6x6+1)2
f '(x)=[1 -30x6]/(6x6+1)2
f '(x)<0
[1 -30x6]/(6x6+1)2<0
1 -30x6<0
30x6>1
x>1/630
terms on series are nonincreasing.f '(x)=[1 -30x6]/(6x6+1)2 <0 for x>1/630, the terms of ak decrease for k>1
limk->ak
=limk->k/(6k6+1)
=limk->1/(6k5+(1/k))
=1/(+0)
=1/
=0
the series converges
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.