M413notes9 - 9 Limit Theorems for Sequences The theorems of...

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9 Limit Theorems for Sequences The theorems of this section allow complicated limits to be evaluated more easily. Theorem. Suppose ( x n ) is a convergent sequence and suppose ( y n ) is a convergent se- quence. Then the sequence ( x n + y n ) is convergent. Moreover, if x n a and y n b then ( x n + y n ) ( a + b ). proof. The basic idea is that if x n is close to a , and y n is close to b , then x n + y n ought to be close to a + b . Let ±> 0. Then there is N 1 such that n>N 1 implies | x n - a | <±/ 2. There is also N 2 such that n>N 2 implies | y n - b | <±/ 2. Let N = max { N 1 ,N 2 } . So if n>N , then n>N 1 and n>N 2 . It follows that | x n - a | <±/ 2 and | y n - b | <±/ 2. So if n>N , we have | ( x n + y n ) - ( a + b ) | = | x n - a + y n - b | ≤| x n - a | + | y n - b |≤ ±/ 2+ ±/ 2= ± so we are done. We would like to prove the similar theorem concerning multiplication, but we need to know about bounded sequences Frst. DeFnition. We say ( x n )is bounded if there is a number M such that | x n |≤ M for all n . We call M a bound for the sequence. If the sequence is not bounded, we say it is unbounded . That is, for any M , there is n such that | x n | >M . Examples. x n = 2 n n +1 is bounded ( M = 2 works). The sequence x n =ln n is unbounded. ±or example, if we let M = 100, we can Fnd n such that ln n>M , i.e., ln n> 100. It turns out n =3 × 10 43 works. Theorem. If ( x n ) is convergent, it must be bounded. proof. Suppose x n L . Let ± = 1. There is N such that n>N implies | x n - L | =1. So for n>N , we must have | x n |≤| L | + 1. (Use || x |-| y || ≤ | x - y | .) So if we let M = | L | + 1, we have a bound for x n , for n>N . But it’s possible that | x k | > | L | + 1 for some k between 1 and N . So we let M = max {| x 1 | , | x 2
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M413notes9 - 9 Limit Theorems for Sequences The theorems of...

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