Sequences....how to proove Decreasing/Increasing

[ku1005]

With the following question, I am able to determine the upper/lower bounds, however im not quite sure how to calculate/show that the sequence is decreasing. Usually I would use either the fact that if decreasing

then

a(n)- a(n+1)>0

or

a(n+1)/a(n)<1

I have included all my relevent workings, any tips on prooving whether increasing/decreasing for further inductively defined sequences would be greatly appreciated, becasue these sorts of problems seem to be easily answered with the "Monotone Sequence Theorem", which once upper/lower bounds determined and whetehr increasing/decreasing....problem becomes easy.


Define:




"Show that 1<=a(n)<=2 for all n>=1 and that the sequene is decreasing"


to show upper/lower bounds I did



Now my attempt at showing decreasing is as follows, but i dont think its correct.





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[Opalg]

That argument looks completely correct to me. Why are you unhappy with it?

 

[ku1005]

I dont agree with what I do for the last step (in proving that it is deacreasing)...but it works????....which I dont like.



ie

a(n) is bound above by 2 and below by 1, this means however that, at some stage, 2 is divided by something at least very clost to 1 if not 1, and therfore is clearly greater then 2/2 = 1 as opposed to < 1 as required.

[ie for my argument to work here, a(n)> 2 for all values of n]

Therefore, Im simply asking someone to show me how to prove the sequence is decreasing, and I included my workings to show you where my train of thought is at. Thanks for your help though.

 

[Opalg]

That has made me think again about this problem, and I noticed something that I should have seen before.

If \(\displaystyle a_1=1\) and \(\displaystyle a_{n+1} = 2/(3-a_n)\) then \(\displaystyle a_2=2/(3-1)=1\), and so (by induction) a_n=1 for all n. So the sequence is in fact constant.

 

[ku1005]

bugger...i didnt mean that at all to happen...and I hadnt realised it myself!!!...so thanks...