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PROBLEM 3. [2]a) Suppose (an)n?? and (bn)n?? are two bounded sequences in ? and prove that lim sup(an + bn) ? lim sup an + lim sup bn (this is problem 12.4 in the text and there is a hint there). [3]b) Give two sequences (an) and (bn) such that lim sup(an + bn) < lim sup an + lim sup bn.

          PROBLEM 3. [2]a) Suppose (an)n?? and (bn)n?? are two bounded sequences in ? and prove that lim sup(an + bn) ? lim sup an + lim sup bn (this is problem 12.4 in the text and there is a hint there). [3]b) Give two sequences (an) and (bn) such that lim sup(an + bn) < lim sup an + lim sup bn.
        
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PROBLEM 3. [2]a) Suppose (an)n?? and (bn)n?? are two bounded sequences in ? and prove that lim sup(an + bn) ? lim sup an + lim sup bn (this is problem 12.4 in the text and there is a hint there). [3]b) Give two sequences (an) and (bn) such that lim sup(an + bn) < lim sup an + lim sup bn.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Suppose (an) n ∈ ℕ and (bn) n ∈ ℕ are two bounded sequences in ℑ and prove that lim sup(an + bn) ≤ lim sup an + lim sup bn (this is problem 12.4 in the text and there is a hint there). Give two sequences (an) and (bn) such that lim sup(an + bn) < lim sup an + lim sup bn.
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Transcript

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00:01 In this question, we have been given that a .n and b are two bounded sequences.
00:05 So a .n and bn, they are given to be the bounded sequences in r.
00:14 Okay.
00:16 We need to prove the following that limit supremum of a .n plus bn is always less or equals to the limit supremum of a .n.
00:31 Limit supremum of bn okay let us see how we are going to do this i will call so let a k to be equals to the supremum of a n such that n is greater or equals to k and i will take another one which is bk which is the supremum of bn such that n is greater or equals to k now we define that limit supremum of a as n tending to infinity it will be nothing but limit k tending to infinity times of a k and similar way limit supremum of b n as n tending to infinity it is nothing but limit k tending to infinity of bk okay so consider k k to be equals to the supremum of sum of a n and b n for all and greater are equals to k then in that case if you try to find what is limit k tending to infinity k it will be limit n tending to infinity supremum of a n plus bn this is what we will get correct now we need to find an index k such that then what we can say for every n which is greater or equals to k we can write that a n plus bn will be less or equals to a k plus bk okay because we estimate that supremum of supremum of all the a n's okay with n greater or equals to k correct now what do we can say ck is equal to supremum of a n plus b n this is true for every n greater or equals to k and it will be less or equals to a k plus b times of k and this holds this will hold for all the value of k now what i will do i will take the limit so limit n tending to infinity of supremum of a n plus bn is always going to be less than supremum of sorry here i will write limit n tending to infinity supremav of a n plus limit and tending to infinity supremum of bn this is what we needed to prove in the next part we need to give an example okay so let us see the next part we need to find two sequences a and bn which satisfies the above condition that is the limit supremum of a .n plus bn should be less than limit supremum of a .n plus limit supremum of bn which you need to give two such examples...
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