This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MTH 122 — Calculus II Essex County College — Division of Mathematics and Physics 1 Lecture Notes #18 — Sakai Web Project Material 1 Introduction to Sequences and Series, Part V 1. The comparison test that we used prior, relies on verifying an inequality between a n and b n , however difficult this may be. To avoid this, we can instead use the following test. Limit Comparison Test : Suppose that ∞ X n =1 a n and ∞ X n =1 b n are series with positive terms. If lim n →∞ a n b n = c where c > 0 is a finite positive number, then either both series converge or both series diverge. Use the limit comparison test to determine if the following series converge or diverge. (a) ∞ X n =1 n 2 + 6 n 4 2 n + 3 Work: Let a n = n 2 + 6 n 4 2 n + 3 , and since a n behaves like 1 /n 2 (a convergent pseries) as n → ∞ , let b n = 1 n 2 . We have lim n →∞ a n b n = lim n →∞ n 4 + 6 n 2 n 4 2 n + 3 = 1 . The limit comparison test applies with c = 1. Since the pseries converges, and c = 1 > 0, this test shows that ∞ X n =1 n 2 + 6 n 4 2 n + 3 converges . (b) ∞ X n =1 sin 1 n Work: You may recall that sin x ≈ x for x near zero. So as n → ∞ the sin 1 n ≈ 1 n . Let a n = sin 1 n , and since a n behaves like 1 /n (a divergent pseries, called the harmonic series) as n → ∞ , let b n = 1 n . We have (this limit was done geometrically....
View
Full Document
 Spring '10
 Ban
 Calculus, Division, Harmonic Series, Sequences And Series, Mathematical Series, lim, n=1

Click to edit the document details