No I am not and I just spotted an error you did too (I think).
=
since is larger then 1 then by "induction"? Correct or something missing?[/quote] this is wrong---
It should be:
And since can only be larger then 1 by induction ?? (I think)
Well, I thank tonio and you for your help, but Iám sorry I do not fully understand. That "quote" wasnt suppose to be there.
So far so god.
Now I have to show that the upper bound is 2, and that is where I fail.
How do I know the upperbound?
Tonio did
How do I solve find this It must be 2, but I cant prove that..
By induction all and the but I dont know how to prove it.
You dont mean that anyone is suppose to do the exercise for me, but I would prefer explanation by baby steps...
Henryt999 ,first of all you must write down a few terms of the sequence to get a "feeling" of what the sequence look like.
And we have that:
.
From the above we have that: .
So we see that the sequence is positive and increasing ,hence can we prove in general that :
1) ?
2) ?
The first one we can prove by induction .
For the 2nd one we must prove: ,or ,since we have proved that .
...............or................................. ...........
.
For that you must prove that: for all nεN ,since for all nεN.
So far having proved that the sequence is psitive increasing and bounded from above by 2 ,the sequence definitely converges to a limit x>0
Now if a sequence converges to x ,every subsequence of that sequence converges to x ,hence :
,since is a subsequence of .
Thus .
AND x=-1 or x=2
I feel that your second statement is much too complicated. It is possible to show that for all via induction, as well.So we see that the sequence is positive and increasing ,hence can we prove in general that :
1) ?
2) ?
The first one we can prove by induction .
For the 2nd one we must prove: ,or ,since we have proved that .
You have already checked some base cases, so suppose that . Now, add two to each side. . Now take the square root of both sides, keeping in mind that these terms are positive and square root is monotonic. . But, by definition, and . So we have . This fulfills the induction step and hence, monotonicity is proved.