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Unformatted text preview: Ma1a 2007 Exam 1 Solutions Question 1. Prove that for n = 2 , 3 ,... , 1 1 4 1 1 9 Â·Â·Â· 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 â‰¥ 2 . Our base case is A (2), which is easily verified: 1 1 2 2 = 3 4 = 2 + 1 2 Â· 2 . Now, assuming that A ( n ) is true for some n â‰¥ 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 Â· 1 1 ( n + 1) 2 = n + 1 2 n Â· ( n + 1) 2 1 ( n + 1) 2 ! = n + 1 2 n Â· 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 â‰¥ 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....
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 Fall '08
 Borodin,A
 Calculus, Linear Algebra, Algebra, 2 K, 1 K, âˆž

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