Wednesday, 27 April 2016

real analysis - Evaluate the limit using only the following results





Let u2>u1>0 and also let un+1=unun1 for all n2. Then prove that the sequence {un} converges.




For this, use of only the following results is permissible,




  1. Archimedean Property.


  2. For two sequence {xn} and {yn}





limn(xn+yn)=limnxn+limnyn



limn(xnyn)=(limnxn)(limnyn)



provided they exists.




  1. For a sequence {zn}, the limit limnzn doesn't exist is equivalent to saying that, ε>0|znl|ε lRnn0(N)




I have been able to prove that limn(unu2n+1)=u1u22. Now one can conclude from 1 and 2 that the limit must be 3u1u22 but that happens only if we can prove that the limits exist. And this is exactly where I am stuck. Using only the three mentioned results I can't prove that. Any help will be appreciated.


Answer



This is similar to what you was trying, but if it helps... (apologies for my poor English)



Supposing you are allowed to use induction, then
u2n+1=unun1u2n+1un=u2nun1=...=u22u1



Making u22u1=a you have un+1=a/un=4aun1 and again by induction un+1=asnucni

i{1,2}, sn is a partial sum of a geometric series, and cn a power of 14, all depending on the parity of n.



Now you can use the result number (3) to prove that each element on the right member of the last equality has a limit (I think this is an easier problem than the initial), and then use result (2) to conclude.


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