SalasSV_10_03 - 10.3 LIMIT OF A SEQUENCE You have seen the...

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Unformatted text preview: 10.3 LIMIT OF A SEQUENCE You have seen the limit process applied in various settings. The limit process applied to sequences is exactly what you would expect. DEFINITION 10.3.1 LIMIT OF A SEQUENCE lim n a n = L if for each > 0, there exists a positive integer K such that if n K , then | a n L | < . Example 1 lim n 4 n 1 n = 4. PROOF Let > 0. We must show that there exists an integer K such that if n K , then 4 n 1 n 4 < . Note that 4 n 1 n 4 = 4 n 1 4 n n = 1 n = 1 n . 596 CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS Choose an integer K > 1 / . Then 1 / K < . Now, if n K , then 1 / n 1 / K and 4 n 1 n 4 = 1 n 1 K < . Example 2 lim n 2 n n + 1 = 2. PROOF Let > 0. We want a positive integer K such that if n K , then 2 n n + 1 2 < . Note that 2 n n + 1 2 = 2 n 2( n + 1) n + 1 = 2 n + 1 = 2 n + 1 < 2 n . We want 2 / n < . We can get this by having n > 2 / ; that is, by having n > (2 / ) 2 . Choose K > (2 / ) 2 . Then n K implies n > (2 / ) 2 , which forces 2 n n + 1 2 < . The next example justifies the familiar statement 1 3 = 0. 333 ... . Example 3 The decimal fractions a n = n 0. 33 ... 3, n = 1, 2, 3, ... tend to 1 3 as a limit: lim n a n = 1 3 . PROOF Let > 0. In the first place (1) a n 1 3 = n 0. 33 3 1 3 = n 0. 99 9 1 3 = 1 3 1 10 n < 1 10 n . Now choose K so that 1 / 10 K < . If n K , then by (1) a n 1 3 < 1 10 n 1 10 K < . Limit Theorems The limit process for sequences is so similar to the limit processes you have already studied that you may find you can prove many of the limit theorems yourself. In any case, try to come up with your own proofs and refer to these only if necessary. 10.3 LIMIT OF A SEQUENCE 597 THEOREM 10.3.2 UNIQUENESS OF LIMIT If lim n a n = L and lim n a n = M , then L = M . A proof, similar to the proof of Theorem 2.3.1, is given in the supplement at the end of this section. DEFINITION 10.3.3 A sequence that has a limit is said to be convergent . A sequence that has no limit is said to be divergent . Instead of writing lim n a n = L , we will often write a n L (read a n converges to L ) or more fully, a n L as n . THEOREM 10.3.4 Every convergent sequence is bounded. PROOF Assume that a n L and choose any positive number: 1, for instance. Using 1 as , you can see that there must exist a positive integer K such that | a n L | < 1 for all n K . Since | a n | | L | | a n | | L | | a n L | , we have | a n | < 1 + | L | for all n K ....
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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SalasSV_10_03 - 10.3 LIMIT OF A SEQUENCE You have seen the...

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