Unformatted text preview: ≤ b n frequently. If a > b , then a-b > 0 and a n-b n → a-b implies by the ﬁrst part that ultimately ( a n-b n ) > 0, which contradicts a n ≤ b n frequently. Page 41 Problem 7. Observe that a n +1-a n = 1 2 n +1 + 1 2( n +1)-1 n +1 = 1 2 n +1-1 2( n +1) > 0, so ( a n ) is increasing. To bound a n from above, note that a n is a sum of n terms each of which is less or equal to 1 n +1 , so a n ≤ n n +1 < 1. Hence ( a n ) is increasing and bounded above, and thus convergent (its limit is the supremum). Page 47 Problem 3. True, since | sin n | ≤ 1 for all n , we have a bounded sequence, which by the Bolzano-Weierstrass theorem has a convergent subsequence. 1...
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- Fall '10
- Mathematical analysis, Limit superior and limit inferior, Xn