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
Unformatted text preview: Ma1a 2007 Exam 1 Solutions Question 1. Prove that for n = 2 , 3 ,... , 1 1 4 1 1 9 1 1 n 2 = n + 1 2 n . Solution 1. Let A ( n ) be the statement that n Y k =2 1 1 k 2 = n + 1 2 n . We will prove this statement is true for n 2 . Our base case is A (2), which is easily verified: 1 1 2 2 = 3 4 = 2 + 1 2 2 . Now, assuming that A ( n ) is true for some n 2, we will prove that A ( n +1) is true as well. Thus, beginning with the left hand side of A ( n +1), we have n +1 Y k =2 1 1 k 2 = n Y k =2 1 1 k 2 1 1 ( n + 1) 2 = n + 1 2 n ( n + 1) 2 1 ( n + 1) 2 ! = n + 1 2 n n ( n + 2) ( n + 1) 2 = ( n + 1) + 1 2( n + 1) . Of course, this last term is the righthand side of A ( n + 1). By induction, we conclude that A ( n ) is true for all n 2. Question 2. (a) Show that if a monotonically increasing sequence is bounded from above then it has a limit. (b) Give an example of a sequence which is bounded from above and from below, and which does not have a limit. (You need to prove the nonexis tence.) Solution 2....
View Full
Document
 Fall '08
 Borodin,A
 Calculus, Linear Algebra, Algebra

Click to edit the document details