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Advanced Math/Real Analysis Proofs

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Expert: - 3/3/2009


Question
Hi again, I have some proofs that I'm having trouble with and would greatly appreciate your help!

1)  Consider three sequences, (a subscript n) (b subscript n) and (c subscript n).  Suppose that (a subscript n)<= (b subscript n) <= (c subscript n) for all n in the set of Natural Numbers.  Suppose that sequence (a subscript n) converges to L and sequence (c subscript n) converges to L. Prove that (b subscript n) converges to L also.

2.  Suppose that a is a number and that (a subscript n) is a bounded sequence such that every convergent subsequence of (a subscript n) converges to a. Prove that (a subscript n) converges to a.

3.  Prove that since 2 subscript x sub n < 2 subscript x sub n+1, it follows that the sqare root of (2 subscript x sub n) < the square root of (2 subscript x sub n+1).  Either prove this or refute the statement.  

Answer
1.  If Bn is less than An and greater than Cn, and both An and Cn converge to some point, then when we're given a value E1 for the error such that A is closer than E1 for any greater n and E2 for the error in C.  Since we know thatt Bn lies between An and Cn,
we know that the distance of it away from the convergent point is less than E1.

2. Suppose there was some subsequence that didn't convrege to that point A.  Then define it's terms as a new subsequence.  But it states that all subsequences convege to the point A, so this is not the case.

3.  Take some An where √An > √An+1.  Take E as the difference,
so √An + E = √An+1.

Square both sides any you get An + 2(√An)E + EČ = An+1.
E was positive, so EČ is positive and 2(√An)E is positive.  Using this, we can say that An+1 is greater that An, which we know is not true.

Therefore we can find no √An > √An+1.

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