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SEQUENCES
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It should be emphasized that the use of the notations 1 and 1 for limits does not in any way imply convergence of the given sequences, since 1 and 1 are not numbers. Instead, these are notations used to describe that the sequences diverge in speci c ways.
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2.7. Prove that lim xn 0 if jxj < 1.
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Method 1: We can restrict ourselves to x 6 0, since if x 0, the result is clearly true. Given  > 0, we must show that there exists N such that jxn j <  for n > N. Now jxn j jxjn <  when n log10 jxj < log10 . Dividing by log10  N, proving the required result. log10 jxj, which is negative, yields n > log10 jxj Method 2: Let jxj 1= 1 p , where p > 0. By Bernoulli s inequality (Problem 1.31, 1), we have jxn j jxjn 1= 1 p n < 1= 1 np <  for all n > N. Thus lim xn 0.
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THEOREMS ON LIMITS OF SEQUENCES 2.8. Prove that if lim un exists, it must be unique.
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We must show that if lim un l1 and lim un l2 , then l1 l2 .
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n!1 n!1
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By hypothesis, given any  > 0 we can nd N such that jun l1 j < 1  when n > N; 2 Then jl1 l2 j jl1 un un l2 j @ jl1 un j jun l2 j < 1  1   2 2 i.e., jl1 l2 j is less than any positive  (however small) and so must be zero. Thus, l1 l2 . jun l2 j < 1  when n > N 2
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2.9. If lim an A and lim bn B, prove that lim an bn A B.
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n!1 n!1 n!1
We must show that for any  > 0, we can nd N > 0 such that j an bn A B j <  for all n > N. From absolute value property 2, Page 3 we have j an bn A B j j an A bn B j @ jan Aj jbn Bj By hypothesis, given  > 0 we can nd N1 and N2 such that jan Aj < 1  2 jbn Bj < 1  2 Then from (1), (2), and (3), j an bn A B j < 1  1   2 2 where N is chosen as the larger of N1 and N2 . for all n > N for all n > N1 for all n > N2 2 3 1
Thus, the required result follows.
2.10. Prove that a convergent sequence is bounded.
Given lim an A, we must show that there exists a positive number P such that jan j < P for all n. Now
jan j jan A Aj @ jan Aj jAj But by hypothesis we can nd N such that jan Aj <  for all n > N, i.e., jan j <  jAj for all n > N It follows that jan j < P for all n if we choose P as the largest one of the numbers a1 ; a2 ; . . . ; aN ,  jAj.
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SEQUENCES
2.11. If lim bn B 6 0, prove there exists a number N such that jbn j > 1 jBj for all n > N. 2
Since B B bn bn , we have: (1) jBj @ jB bn j jbn j. Now we can choose N so that jB bn j jbn Bj < 1 jBj for all n > N, since lim bn B by hypothesis. 2 n!1 Hence, from (1), jBj < 1 jBj jbn j or jbn j > 1 jBj for all n > N. 2 2
2.12. If lim an A and lim bn B, prove that lim an bn AB.
n!1 n!1 n!1
We have, using Problem 2.10, jan bn ABj jan bn B B an A j @ jan jjbn Bj jBjjan Aj @ Pjbn Bj jBj 1 jan Aj But since lim an A and lim bn B, given any  > 0 we can nd N1 and N2 such that
 for all n > N1 jbn Bj < 2P
jan Aj <
 for all n > N2 2 jBj 1
Hence, from (1), jan bn ABj < 1  1   for all n > N, where N is the larger of N1 and N2 . Thus, the 2 2 result is proved.
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