mth.122.handout.18

# mth.122.handout.18 - MTH 122 — Calculus II Essex County...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the 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 p-series) 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 p-series 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 p-series, called the harmonic series) as n → ∞ , let b n = 1 n . We have (this limit was done geometrically....
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

mth.122.handout.18 - MTH 122 — Calculus II Essex County...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online