Thursday, March 21, 2019

real analysis - Limit of the sequence given by xn+1=xnxn+1n



Let , x1(0,1) be a real number. For n>1 define xn+1=xnxn+1n. Then prove that limnxn exists.



We have to prove that the given sequence {xn} is convergent. So we have to show that {xn} is monotone and bounded.


I proved that the sequence is monotone decreasing. But I'm unable to show that it is bounded below. How can I show it ?



Any other way to prove that the limit exists ?


Answer



We show by induction that xn(0,1) for all n:


The case n=1 is clear.


Now let nN and xn(0,1)


Then: xn+1=xn(1xnn). From xn(0,1) we get xnn(0,1) and therefore 1xnn(0,1).


Consequence: xn+1(0,1).


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