- #1

mattmns

- 1,128

- 6

Hello, here is the exercise from the book:

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Let [itex](a_{n})_{n=0}^{\infty}[/itex] be a sequence which is not bounded. Show that there exists a subsequence [itex](b_{n})_{n=0}^{\infty}[/itex] of [itex](a_{n})_{n=0}^{\infty}[/itex] such that the limit [itex]\lim_{n\rightarrow \infty} 1/b_{n} = 0[/itex]. (Hint: for each natural number j, introduce the quantity [itex]n_{j} := min\{n \in \mathbb{N} : |a_{n}| \geq j \}[/itex], first by explaining why the set [itex]\{n \in \mathbb{N} : |a_{n}| \geq j \}[/itex] is non-empty. Then set [itex]b_{j} := a_{n_{j}}[/itex].)

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First the hint seems incorrect to me. What I mean is that we could have a situation with b1 = b2 = b3 = a1. And have only one element in the sequence with the value of a1. Consider the following: An = -4,-3,-2,-1,0,1,2,3,4,... This sequence is not bounded (a sequence {An} is bounded by a real number M iff |An| <= M for all n.) so it meets the requirement. But when we look at n0, n1, n2, we get the same value, 0. Thus our sequence bn would start out: -4,-4,-4,-4, then the rest, but this is not necessarily a subsequence of An. Am I missing something here, or is the hint incorrect?

Regardless of that I think I have the idea behind a (the) solution. First let us pick b1 as the first element in An that is larger than 1. Then we will pick b2 as the next element in An that is larger than b1, which we can do since An is not bounded, and then continue. This will give us a sequence that is strictly increasing. Then when we look at 1/bn we will have a sequence that is strictly decreasing, but more importantly we will have [itex]0 \leq 1/b_{n} \leq 1/n[/itex] so that when we take the limit we can use the squeeze test to get [itex]\lim_{n\rightarrow \infty} 1/b_{n} = 0[/itex]. My problem with this is that I do not know how to formalize this, which is what I think our professor wants, and also I would like to know how to do such a thing. What I mean by formalize is give a more explicit definition of what bn is. Any ideas?

Thanks!

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Let [itex](a_{n})_{n=0}^{\infty}[/itex] be a sequence which is not bounded. Show that there exists a subsequence [itex](b_{n})_{n=0}^{\infty}[/itex] of [itex](a_{n})_{n=0}^{\infty}[/itex] such that the limit [itex]\lim_{n\rightarrow \infty} 1/b_{n} = 0[/itex]. (Hint: for each natural number j, introduce the quantity [itex]n_{j} := min\{n \in \mathbb{N} : |a_{n}| \geq j \}[/itex], first by explaining why the set [itex]\{n \in \mathbb{N} : |a_{n}| \geq j \}[/itex] is non-empty. Then set [itex]b_{j} := a_{n_{j}}[/itex].)

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First the hint seems incorrect to me. What I mean is that we could have a situation with b1 = b2 = b3 = a1. And have only one element in the sequence with the value of a1. Consider the following: An = -4,-3,-2,-1,0,1,2,3,4,... This sequence is not bounded (a sequence {An} is bounded by a real number M iff |An| <= M for all n.) so it meets the requirement. But when we look at n0, n1, n2, we get the same value, 0. Thus our sequence bn would start out: -4,-4,-4,-4, then the rest, but this is not necessarily a subsequence of An. Am I missing something here, or is the hint incorrect?

Regardless of that I think I have the idea behind a (the) solution. First let us pick b1 as the first element in An that is larger than 1. Then we will pick b2 as the next element in An that is larger than b1, which we can do since An is not bounded, and then continue. This will give us a sequence that is strictly increasing. Then when we look at 1/bn we will have a sequence that is strictly decreasing, but more importantly we will have [itex]0 \leq 1/b_{n} \leq 1/n[/itex] so that when we take the limit we can use the squeeze test to get [itex]\lim_{n\rightarrow \infty} 1/b_{n} = 0[/itex]. My problem with this is that I do not know how to formalize this, which is what I think our professor wants, and also I would like to know how to do such a thing. What I mean by formalize is give a more explicit definition of what bn is. Any ideas?

Thanks!

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