Constant sigh-how can I show it??

evinda

Junior Member
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Apr 13, 2013
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57
Hi!
Given that \(\displaystyle a_{1}>0\) and \(\displaystyle a_{n+1}=1+\frac{2}{1+a_{n}}\). I am asked to prove that the two subsequences \(\displaystyle a_{2k}\) and \(\displaystyle a_{2k-1}\) are monotone. To show this I have to show that the difference \(\displaystyle a_{n+2}-a_{n} \) has constant sign, but how can do that?
I hope someone can help me..
 
Write \(\displaystyle a_{n+2}\) in terms of \(\displaystyle a_n\)
 
Write \(\displaystyle a_{n+2}\) in terms of \(\displaystyle a_n\)

I wrote this, and I got \(\displaystyle a_{n+2}-a_{n}=\frac{3-a_{n}^2}{2+a_{n}} \), how can show that the sign is constant?
 
I wrote this, and I got \(\displaystyle a_{n+2}-a_{n}=\frac{3-a_{n}^2}{2+a_{n}} \), how can show that the sign is constant?

We have \(\displaystyle a_{n+2} = 2-\dfrac{1}{2+a_n}\)

Case 1: \(\displaystyle a_3-a_1>0\). Prove \(\displaystyle a_{2n+1}-a_{2n-1} > 0 \iff a_{2n-1}-a_{2n-3}>0\) and conclude by induction the odd members are increasing.

Then show if \(\displaystyle a_3-a_1 > 0\) then \(\displaystyle a_4-a_2<0\) and \(\displaystyle a_{2n+2}-a_{2n} < 0 \iff a_{2n}-a_{2n-2}>0\). Conclude that the even members are decreasing.

Case 2: \(\displaystyle a_3-a_1<0\), do something similar.

Beyond some algebra it isn't very bad. To get you started:

\(\displaystyle a_5-a_3 > 0 \iff \left[2-\dfrac{1}{2+a_3}\right] -\left[2-\dfrac{1}{2+a_1}\right] >0 \iff \dfrac{a_3-a_1}{(2+a_1)(2+a_3)}>0 \)
 
Could you explain me why when the difference of the even terms is >0 ,the difference of the odd terms is <0 ????
 
Could you explain me why when the difference of the even terms is >0 ,the difference of the odd terms is <0 ????

\(\displaystyle a_5-a_3 > 0 \iff \left[1+\dfrac{2}{1+a_4}\right] - \left[1+\dfrac{2}{1+a_2}\right] >0 \iff \dfrac{2(1+a_2)-2(1+a_4)}{(1+a_4)(1+a_2)}>0\)

see it?
 
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