A question regarding a new solution

[transgalactic]

i recently posted a question about prooving a convergence of a series
and to find the limit

i have found a new way to proove
i showed it in the link

is this method ok??

the problem is in the second step
when i am trying to prove that the series doesn't go bellow 0.61

i used the splitting method
of the fraction but its inconclusive

how do i proove the second part
?

do i need to proove anything else inorder to complete the objective?

http://img134.imageshack.us/my.php?image=img8220ez9.jpg

(there is a typing miste in the link its a n+1>0.61)

 

[sutupidmath]

i recently posted a question about prooving a convergence of a series
and to find the limit

i have found a new way to proove
i showed it in the link

is this method ok??

the problem is in the second step
when i am trying to prove that the series doesn't go bellow 0.61

i used the splitting method
of the fraction but its inconclusive

how do i proove the second part
?

do i need to proove anything else inorder to complete the objective?

http://img134.imageshack.us/my.php?image=img8220ez9.jpg

(there is a typing miste in the link its a n+1>0.61)

This looks like a sequence to me!

 

[sutupidmath]

Why don't u post the original question first?

 

[transgalactic]

the original question and how i tried to solve it in the new link

on the top of the first page

 

[HallsofIvy]

You say, at one point, "$a_k> 0.61$ given". Well, you aren't "given" that. And I don't see where you have proved it. But then you say "we need to prove that a_k+1< 0.61" which isn't true. Are you saying that a_k> 0.61 for some k? Well, a_k+1 < 0.061 is never true, any way.

I think it is simpler to prove the contrapositive. Suppose that, for some k, a_k+1> a_k. Then
[tex]a_{k+1}= \frac{a_k+ 1}{a_k+ 2}> a_k[/itex]
Multiply on both sides by the positive value $a_k+ 2$. Then you are saying that $a_k+ 1> a_k^2+ 2a_k$ so that $a_k^2+ a_k-1< 0$. That's a parabola opening upward. For what values of x is x²+ x- 1< 0? Are those possible values of a_k?

 

[transgalactic]

the parabula shows how the series goes

if the line after some point goes is on the negative part of the y axes
(bellow the x axes) then each next member will decrease in value

thats how i overrided the prooving
of

an+1<an inequality

you showed yourself the parula formula in the numanator
so i think it is the same.
is it ok??

now i want to proove that
its not going bellow 0.61

by the induction method

we presume that
an>0.61

and using that we proove that

an+1>0.61

as i showed in the link
i am having trouble to solve it
??

 

[Rainbow Child]

Since $a_0=1$ then

$$a_{k+1}= \frac{a_k+ 1}{a_k+ 2}> 0$$

by induction, i.e. $a_k>0 \,\forall\, k$. Thus the limit is

$$l=\frac{-1+\sqrt{5}}{2}$$