ddddd Use the Comparison Test or the Limit Comparison Test to determine whether
ID: 2848424 • Letter: D
Question
ddddd
Use the Comparison Test or the Limit Comparison Test to determine whether the series converges or diverges. The Comparison Test - Reminder: Suppose that and are series with positive terms. If is convergent and for all n, then, is also convergent. If is divergent and for all n, then is also divergent. The Limit Comparison Test - Reminder: Suppose; that an and are series with positive terms. If 11 111 = C n a n where c is a finite number and 0, then either both series converge or both diverge. the series The Comparison Test - Reminder: Suppose that an and 5Z are series with positive terms. (i) If is convergent and for all n, then is also convergent. If n Ls divergent and bn for all is also divergent. Estimate for the Integral Test - Remainder: Suppose f(i) = a,, where f Is continuous, positive, decreasing function for n and a is convergent with s = Estimate for the Integral Test-Remainder: Suppose where f is continuous, positive, decreasing function for is convergent with s = where sn = then Solution :since Choices convergent, divergent by the Integral Test, convergence Test, Remainder Estimate, Comparison Test.Explanation / Answer
1)
a) Using limit test with bn=1/n the limit of an/bn = 10^10 , since bn series diverges , then an diverges
b) Same than in a) limit an/bn = 1 with bn=1/n so it diverges
c) with bn=1/n^2 this time , limit of an/bn = 1 and since bn converges, then an converges ( p-test with p=2 > 1) or integral test
d) by comparison this time 1/(1.1^n + 4) <= 1/1.1^n = (10/11)^n
The latter converges since 10/11 < 1 so an converges.
e) with bn=1/n the limit of an/bn = pi/2 and bn diverges, so an diverges.
2)
a) 1/(3i^3 + 10) <= 1/(3i^3) so choice is < and 1/(3i^3)
It's convergent by comparison test
b) int 1/(3x^3) = -1/(6x^2) so the answer is = 1/(6n^2) for limits asked.
c) Rn <= Tn <= 1/(6n^2)
d) R4 <= 1/(6*4^2) by plugging n=4 in previous inequality
e) I,find with computer the sum is 0.1222 and the error is less than 1/(6*4^2)=0.0104
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