# Real Analysis Question

label Mathematics
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schedule 1 Day
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May 2nd, 2015

1.an=1/n is bounded, doesn't have minimum and maximum value in its sequence

an is bounded and therefore, it has an infimum and a supremum. Since it does not have a maximum and minimum, it has infinitely many elements near both infimum and supremum. This means that there are 2 subsequences of an; one of them converges to the infimum and the other converges to the supremum. But infimum and supremum are distinct (if they were not distinct it would be a constant sequence

2.For example, I have this sequence:

A1=a,(a>0,a∈R),An+1=An+(An)2/(1+An)2

It is increasing, hence all terms are ≥a. The function f:x→x+x2/(1+x2) is continuous on (a,∞) and has no fixed points.

Assume that the sequence is bounded. Then it is convergent. The limit is a fixed point of f. You get a contradiction.

....................................post the rest as a 2 hour question...please!!

May 2nd, 2015

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May 2nd, 2015
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May 2nd, 2015
Nov 22nd, 2017
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