{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solutionshw4-554-2010

# solutionshw4-554-2010 - Solutions homework 4 Page 22...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions homework 4. Page 22 Problem 1. : By 1.3.8- a n ↑ - a and- b n ↑ - b , so by Theorem 2.5.6 we have- a n + (- b n ) ↑ - a + (- b ), which again by 1.3.8 implies that a n + b n ↓ a + b . Page 22 Problem 2. From 0 ≤ a n ≤ a n +1 and 0 ≤ b n ≤ b n +1 it follows that a n b n ≤ a n +1 b n +1 , so ( a n b n ) is increasing. Also 0 ≤ a n ≤ a and 0 ≤ b n ≤ b implies that a n b n ≤ ab . It remains to show that ab is the least upper bound of a n b n . To see this we have to consider first the case that a n = 0 for all n ≥ 1. In this case a = 0 and a n b n = 0, so 0 = ab = sup a n b n in this case. Similarly we can deal with the case the case that b n = 0 for all n ≥ 1. We can therefore assume assume that a n 6 = 0 from some index on and b n 6 = 0 from some index on. In particular this implies that a 6 = 0 and b 6 = 0. Let now s ≥ a n b n . Then s b n ≥ a n from some index on, which proves that s b n ≥ a from some index on. Hence s a ≥ b n , which implies...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online