1. 2. (1pt)Usethelimitcomparisontesttodeterminewhether ,,- 4n3+2n2+3 (1 pt) Use

Question

1.

2.

(1pt)Usethelimitcomparisontesttodeterminewhether ,,- 4n3+2n2+3 (1 pt) Use the limit comparison test to determine whether an- converges or diverges. 4n3 2n2 3 72 (a) Choose a series bn with terms of the form bn and apply the limit comparison test. Write your answer as a fully reduced fraction. For n > 4 nP 72 Ca 72 (b) Evaluate the limit in the previous part. Enter OO as infinity and-OO as-infinity. If the limit does not exist, enter DNE. Ca 72 72 (c) By the limit comparison test, does the series converge, diverge, or is the test inconce? Choose

Explanation / Answer

1)an=(9n+7)/(4n3+2n2+3)

bn =1/n2

a)lim n-> an/bn=lim n-> [(9n+7)/(4n3+2n2+3)]/(1/n2)

b)lim n-> [n2(9n+7)/(4n3+2n2+3)]

=lim n-> [n3(9+(7/n))/(n3(4+(2/n)+(3/n2))]

=lim n-> [(9+(7/n))/(4+(2/n)+(3/n2))]

=(9+0)/(4+0+0)

=9/4

c)limit is finate and nonzero, and n=4 to 1/n2 converges , therefore by limit comparision test

n=4 to (9n+7)/(4n3+2n2+3) converges

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.