Fill in the blanks in the following proof.Claim: Let x₁ = 1 + for all n E N. Then x, does not converge to zero.nProof: Take e =. ThenNEN we take n =| 1 + ² = 1 +1 + 1 ≥ €which completes the proof.N+11N-1>for allwe choose2Then n0N and we verify:

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Answered: Fill in the blanks in the following…

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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Determine whether converges or diverges. Show complete solution.
25. Let an := n 2 1. Which of these conditions must be met if an is to converge n+1 to 1? I. for any e > 0, n > II. for any e > 0, | II. for any e > 0, | ke vn2 n(e). |> e Vn 2 n(e).. n+: A. I and II only В. I only C. II and III only D. I and III only E None of the above choices A, B, C or D..
n Find a formula for |I (1 and verify. k2 k=2
May you please show all steps.
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In which of the following options, the statements given below and the general terms an that satisfy these statements are matched correctly? I: is divergent when > an is convergent n=1 n=1 II: E- is divergent when >an is divergent n=1 an n=1 is convergent when the sequence {an} an n=1 III: is divergent K: an = n2 - Vn 1 an L: an M: = - 3 n (а) 1-М (b) I-L (c) I-M (d) I-L (е) I-К II - K II -M II - L II - K II - M III - L III -K III - K III - M III - L
(b) Decide whether the following are true or false giving a proof or coun terexample as appropriate. (i) For all a, n, m E Z. If a n and a m then a | (n +m). (ii) For all a, b, n, m E Z. If a | n and b| m then (a + b) | (n + m).
n Let n, m E N with m < n. Find and prove a formula for i. Express your final answer as a simplified i=m fraction involving n, m. (Hint: You should not need induction!)
answer it correctly and show solutions
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Using the same proof that we saw in class that shows that n=177 converges, prove that -1 7a converges when a > 1. 1
Choose the answer that best completes the proof that inf(A)=0 where A = Firstly, we would show that o is a lower bound for A by noticing that 0s for all nEN- n2 Then.. O None of these O Then we would let W>0 be arbitrary but fixed and show that W can't be a lower bound for A: 1 0 it follows, from the Archimedean Property, that there exists NEN such that O Then we would let W>0 be arbitrary but fixed and show that W can't be a lower bound for A: Since W>0 it follows that we can choose EA such that " 0 be arbitrary but fixed and show that W can't be a lower bound for A: Since W>0 it follows, from the Archimedean Property, that there exists NEN such that

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Fill in the blanks in the following proof. Claim: Let x = 1 + for all n E N. Then x, does not converge to zero. n Proof: Take e = . Then NEN we take n = | 1 + ² = 1 + 1 + 1 ≥ € which completes the proof. N+1 1 N-1 > for all we choose 2 Then n 0 N and we verify: