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Lemma 7.2.7. A Convergent Sequence Is Bounded. then there exists B > 0 If limn00 an = a, such that an < B for all n. in-context

Lemma 7.2.7. A Convergent Sequence Is Bounded. then there exists B > 0 If limn00 an = a, such that an < B for all n. in-context

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Analysis math. Problem 7.2.8

10:48 5J Problem 7.2.8. Prove Lemma 7.2.7. Lemma 7.2.7. A Convergent Sequence Is Bounded. If limn0 an = a, then there exists B > 0 such that an| B for all n. in-context Hint. We know that there exists N such that if n > N, then |a, – al < 1. Let B max(la1|, la2|,..., la[|, |a| +1), where |N] represents the smallest integer greater than or equal to N. Also, notice that this is not a convergence proof so it is not safe to think of N as a large number. Actually, this is a dangerous habit to fall into even in convergence proofs. Armed with this bound B, we can add on one more inequality to the above scrapwork to get · b| = |an · bn < lan · bn = |an||bn < B\b, – b| + (|b| + 1) |an |an · bn an ·b+ an · b – an · b| + |an · b – -b| + |6| |a, – a| - II V II
Transcribed Image Text:10:48 5J Problem 7.2.8. Prove Lemma 7.2.7. Lemma 7.2.7. A Convergent Sequence Is Bounded. If limn0 an = a, then there exists B > 0 such that an| B for all n. in-context Hint. We know that there exists N such that if n > N, then |a, – al < 1. Let B max(la1|, la2|,..., la[|, |a| +1), where |N] represents the smallest integer greater than or equal to N. Also, notice that this is not a convergence proof so it is not safe to think of N as a large number. Actually, this is a dangerous habit to fall into even in convergence proofs. Armed with this bound B, we can add on one more inequality to the above scrapwork to get · b| = |an · bn < lan · bn = |an||bn < B\b, – b| + (|b| + 1) |an |an · bn an ·b+ an · b – an · b| + |an · b – -b| + |6| |a, – a| - II V II
10:48 "slick trick." Notice that we can add one more line to the above string of inequalities: |a,||b, – b| + |6||a, – a| < |a,| |b,, - 6| + (|b| + Now we can make |an – a|< 2(6+1) and not worry about dividing by zero. Making |a,||b, – b| < { requires a bit - more finesse. At first glance, one would be tempted to try and make b, – b| < ga: Even if we ignore the fact that we could be dividing by zero (which we could handle), we have a bigger problem. According to the definition of limn-+∞ bn = b, we can make b, – b smaller than any given fixed positive number, as long as we make n large enough (larger than some N which goes with a given epsilon). Unfortunately, is not fixed as it has the variable n in 2|a, it; there is no reason to believe that a single N will work with all of these simultaneously. To handle this impasse, we need the following: Lemma 7.2.7. A Convergent Sequence Is Bounded. If lim, →0∞ An = a, then there exists B > 0 such that |an| < B for all n. II II
Transcribed Image Text:10:48 "slick trick." Notice that we can add one more line to the above string of inequalities: |a,||b, – b| + |6||a, – a| < |a,| |b,, - 6| + (|b| + Now we can make |an – a|< 2(6+1) and not worry about dividing by zero. Making |a,||b, – b| < { requires a bit - more finesse. At first glance, one would be tempted to try and make b, – b| < ga: Even if we ignore the fact that we could be dividing by zero (which we could handle), we have a bigger problem. According to the definition of limn-+∞ bn = b, we can make b, – b smaller than any given fixed positive number, as long as we make n large enough (larger than some N which goes with a given epsilon). Unfortunately, is not fixed as it has the variable n in 2|a, it; there is no reason to believe that a single N will work with all of these simultaneously. To handle this impasse, we need the following: Lemma 7.2.7. A Convergent Sequence Is Bounded. If lim, →0∞ An = a, then there exists B > 0 such that |an| < B for all n. II II
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