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Math 317 hw 11

In other words, I want to show that, if (a n ) is a bounded sequence that does not converge to the number a R, then not every convergent subsequence converges to a (or, more positively, there is some subsequence that converges to b a). (c) A sequence that contains subsequences converging to every point in the infinite set 1, 1/2, 1/3, 1/4, 1/5. Since (a n ) is bounded, the subsequence (a nk ) is also bounded and thus, by the Bolzano-Weierstrass Theorem, contains a convergent subsequence (a nkl ). The sequence (z n ) is increasing and bounded (it s bounded below by z 1 and above by the upper bound for (a n so the Monotone Convergence Theorem implies it converges, so the above limit definitely exists. Exercise Give an example of math 317 hw 11 each of the following, or argue that such a request is impossible. Since our choice of n was arbitrary, we see that a 1 a n M for all n 1, 2, 3,., so the sequence (a n ) must be bounded. On the other hand, y n a n L /2. Since 3, which is an upper bound for (x n we know L Therefore, we can conclude that L Exercise Let (a n ) be a bounded sequence. Provide a reasonable definition for lim inf a n and briefly explain why it always exists for any bounded sequence. Therefore, by induction, we can conclude that 0 x n1 x n 4 for all n 1, 2,. (c) Prove that lim inf a n lim sup a n for every bounded sequence, and give math 317 hw 11 an example of a sequence for which the inequality is strict. Then the limit inferior of (a n ) is defined to be lim inf a n : lim. Then, for any M 0, there exists N N such that N M (by the Archimedean Property then either a N N M or a N1 N. 4 5 (b) A monotone sequence that diverges but has a convergent subsequence.

For all n N, we see that for all 0 there exists N N such that. There is an N N such that. We know that or, for any term a n from the leominster original sequence. If n N 1, which means that a n is a Cauchy sequence. We have 4 x k1 4 x k. Therefore, lesbian since x n converges to, y n L x n1 L since. There exists an N N such that.

Where the first inequality is just an application of the triangle inequality and the second follows from the assumption on n and. Solution 2, show that flowers a n must converge. Decreasing sequence and so, in particular, for any n 2 3 In other words. Eliminating x from the equations of the two planes. A Prove that the sequence defined by y n supa. We have that a n.

If I want to be really rigorous, here s how to do that: define the sequence (y n ) by y n x n1 for all n 1, 2,.Therefore, every element of A is an upper bound for B, so B is bounded above and thus, by the Axiom of Completeness, has a least upper bound supB.On the other hand, the subsequence of odd terms is just the sequence (0, 0, 0,.