Proof Let s n be a convergent sequence and let s be the limit or equivalently s

Proof let s n be a convergent sequence and let s be

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Proof. Let ( s n ) be a convergent sequence, and let s be the limit, or equivalently s = lim s n . Apply the definition of convergence with = 1, then we get some large number N N , such that if n > N , then | s n - s | < 1. By the triangle inequality, | s n - s | < 1 = | s n | ≤ | s + s n - s | ≤ | s | + | s n - s | ≤ | s | + 1 . Define M = max {| s 1 | , | s 2 | , · · · , | s N | , | s | + 1 } . Then we have | s n | ≤ M if n N | s | + 1 M if n > N . Therefore | s n | ≤ M for all n N , and so { s n } is bounded. Remark 9.3 . In the proof above, we only apply the defining condition (6.1) for a single = 1. Actually, we can take to be any finite positive number in the proof, and this is left as an exercise. Theorem 9.4. If s sequence ( s n ) converges to s , and k R , then the sequence ( ks n ) converges to ks . That is, lim( ks n ) = k lim s n . Proof. If k = 0, the proof is trivial as ks n 0, and lim(0 · s n ) = 0 = 0 · lim s n . Assume k 6 = 0 now. Let > 0 be any positive real number, and we need to show that | ks n - ks | < , for large n N . Note that the above inequality is equivalent to saying that | s n - s | < | k | , for large n N .
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24 XIN ZHOU Since lim s n = s , for | k | > 0, by (6.1) there exists N > 0, such that if n N , then | s n - s | < | k | . Therefore for n > N , we have | ks n - ks | < . Theorem 9.5. If ( s n ) converges to s and ( t n ) converges to t , then ( s n + t n ) converges to s + t . That is to say, lim( s n + t n ) = lim s n + lim t n . Proof. Let > 0 be any positive real number, and we want to show that | ( s n + t n ) - ( s + t ) | < , for all large n N . Note that | ( s n + t n ) - ( s + t ) | ≤ | s n - s | + | t n - t | . ( Hint : so we can try to show | s n - s | < / 2 and | t n - t | < / 2.) Since lim s n = s , N 1 N , such that if n N 1 , then | s n - s | < / 2 . Similarly by lim t n = t , N 2 N , such that if n N 2 , then | t n - t | < / 2 . Let N = max { N 1 , N 2 } . Then if n > N , we have | ( s n + t n ) - ( s + t ) | ≤ | s n - s | + | t n - t | < . Theorem 9.6. If ( s n ) converges to s and ( t n ) converges to t , then ( s n · t n ) converges to s · t . That is to say, lim( s n · t n ) = (lim s n ) · (lim t n ) . Discussion : | s n t n - st | = | s n t n - s n t + s n t - st | ≤ | s n || t n - t | + | t || s n - s | M | t n - t | + | t || s n - s | . In the above inequality, | t n - t | and | s n - s | can be very small for large n ; • | s n | is bounded by Theorem 9.2, i.e. | s n | ≤ M .
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MATH 117 FALL 2017 UCSB 25 Proof. Let > 0 be any positive real number. By Theorem 9.2, M > 0, such that | s n | ≤ M , for all n N . Since lim t n = t , N 1 N , such that if n N 1 , then | t n - t | < / (2 M ) . Similarly by lim s n = s , N 2 N , such that if n N 2 , then | s n - s | < 2( | t | + 1) . Let N = max { N 1 , N 2 } . Then if n > N , we have | s n t n - st | ≤ | s n || t n - t | + | t || s n - s | M · 2 M + | t | · 2( | t | + 1) < . Lemma 9.7. If ( s n ) converges to s and if s n 6 = 0 for all n N , and if s 6 = 0 , then lim 1 s n = 1 s . Discussion : 1 s n - 1 s = s n - s s n s | s n - s | m | s | . We know that | s n - s | is very small for n very large. To show the right hand side is small, we need the denominator | s n s | to be bounded away from 0. By Example 7.5, we do have that m = inf {| s n | : n N } > 0 . Proof. Let > 0 be any positive real number, and m be defined as above.
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