Assignment 9 Solutions

Assignment 9 Solutions - MATH 138 Calculus 2 Solutions to...

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

View Full Document Right Arrow Icon
MATH 138 Calculus 2, Solutions to Assignment 9 1: Find the limit of each of the following sequences { a n } , if the limit exists. (a) a n = 4 n 2 + 3 n - n Solution: a n = 4 n 2 + 3 n - n = q 4 + 3 n 2 1 - 1 n 4 1 = 2. (b) a n = ( - 3) n 2 2 n +1 Solution: a n = ( - 3) n 2 2 n +1 = 1 2 ( - 3) n 4 n = 1 2 ( - 3 4 ) n 0. (c) a n = 2 2 n n ! Solution: a n = 2 2 n n ! = 4 n n ! = 4 · 4 · 4 ··· 4 1 · 2 · 3 ··· n = 4 1 · 4 2 · 4 3 · ( 4 4 )( 4 5 )( 4 6 ) ··· ( 4 n - 1 ) · 4 n 4 1 · 4 2 · 4 3 · 4 n = 128 3 1 n 0, since all the terms in brackets are 1. 2: Let a 1 = 4 3 and a n +1 = 5 - 4 a n for n 1. Determine whether { a n } converges, and if so then find the limit. Solution: If { a n } does converge, say a n l , then we also have a n +1 l , and so taking the limits on both sides of the formula a n +1 = 5 - 4 a n gives l = 5 - 4 l = l 2 = 5 l - 4 = l 2 - 5 l + 4 = 0 = ( l - 1)( l - 4) = 0. This shows that if the limit exists then it must be equal to 1 or 4. The first few terms of the sequence are a 1 = 4 3 , a 2 = 2 and a 3 = 3. Since the terms appear to be increasing, we shall try to prove that 1 a n a n +1 4 for all n 1. This is true when n = 1. Suppose it is true when n = k . Then we have 1 a k a k +1 4 =
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

Page1 / 3

Assignment 9 Solutions - MATH 138 Calculus 2 Solutions to...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online