Use the Integral Test to determine whether the series is convergent or divergent
ID: 2882485 • Letter: U
Question
Use the Integral Test to determine whether the series is convergent or divergent. (If the quantity diverges enter DIVERGES.) sigma_n=1^infinity 8/n^4 integral_1^infinity 8/x^4 dx = 2/4 convergent divergent Consider the the following series. Sigma_n=1^infinity 1/n^3 Use the sum of the first 10 terms to estimate the sum of the given series. (Round the answer to six decimal places.) 1.917532 Improve this estimate using the following inequalities with n = 10. (Do this on paper. Your instructor may ask you to turn in this work.) s_n+integral_n+1^infinity f(x) dx lessthanorequalto s lessthanorequalto s_n + integral_n^infinity f(x) dx Find a value of n that will ensure that the error in the approximation is less than 0.001. n > 22Explanation / Answer
we have given summation of n=1 to infinity (8/n^4)
we use function f(x)=8/x4 =8x-4
by using integral test
integration of 8/x4dx from (x=1 to infinity)= limt-->infinity (integration of 8/x4dx from (x=1 to t))
=limt-->infinity (integration of 8x-4dx from (x=1 to t))
=limt-->infinity ((-8/3x^3) from (x=1 to t))
=limt-->infinity ((-8/3t^3)-(-8/3))
=((-8/3)*(1/(infinity)^3)+8/3)
=8/3 since 1/infinity=0
integration of 8/x4dx from (x=1 to infinity) =8/3
the given series is convergent by integral test
a) we have given summation of n=1 to infinity (1/n^3)
let sn=(1/n3)
summation of n=1 to 10 (1/n^3) =1/1+1/8+1/27+1/64+1/125+1/216+1/343+1/512+1/729+1/1000=1.19753198567
summation of n=1 to 10 (1/n^3) =1.197531
b) let f(x)=1/x^3 and we have given n=10
s10=1.197531
integration of (x=11 to infinity) 1/x^3 dx=[(-1)/2x^2] from x=11 to infinity
integration of (x=11 to infinity) 1/x^3 dx= [(-1)/2(infinity)^2+(1)/2*(11)^2]=[0+1/242]=1/242
integration of (x=11 to infinity) 1/x^3 dx=1/242
s10+integration of (x=11 to infinity) 1/x^3 dx=1.197531+1/242=1.2016632314=1.201663
integration of (x=10 to infinity) 1/x^3 dx =[-1/2x^2] from x=10 to infinity
=[(-1/2)*(1/infinity)^2+1/200]=1/200
s10+integration of (x=10 to infinity) 1/x^3 dx=1.197531+1/200=1.202531
s10+integration of (x=11 to infinity) 1/x^3 dx <=s<= s10+integration of (x=10 to infinity) 1/x^3 dx is true
1.201663<=s<=1.202531
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