On a second thought, I think I kind of get it.
If we let e (epsilon) < 1 - L, and a = e + L, e>0, then obviously L<a<1
Please let me know if there is anything wrong with the following thoughts:
Since |s_n+1|<a|s_n| for all n>=N, then |s_N+1|<a|s_N|, and |s_N+2|<a|s_N+1|=a|a|s_N||, so on and...
Finding an L<a<1 isn't difficult, but don't I have to find an N such that n>N implies |S_n+1| < a|s_n|?
From the definition of a limit I got: there exist N s.t. n>N => ||sn+1/sn| - L|< e
which means |sn+1|<(e+L)*|sn|,
I guess we don't need to assign e (epsilon) to be any particular value to...
a negative number... wow I really need to review properties of these functions... thanks
Now the only thing is, how do I find such an 'a'? What epsilon can I choose so that \epsilon + L < 1, when L<1
It seems like any linear combination of L is out of the question, powers of L doesn't seem...
I get log(s_n) < (n-N)log(a) + log(S_N), since n>N as n-> oo, the RHS goes to infinity?! I was hoping that the RHS would go to -oo so that we may conclude |S_n| -> 0
Homework Statement
Assume all sn =/= 0, and that the limit L = lim|(sn+1)/(sn)| exist.
a) Show that if L<1 then lim sn = 0
b) Show that if L>1 then lim sn = + oo
Homework Equations
There is a hint that we should select 'a' s.t. L<a<1 and obtain N s.t. n>N => |sn+1 < a|sn|.
Then show that...
thanks, seems exactly what I needed. To reach that conclusion, can I take the following steps:
Let lim sn = s
Since lim (An + Bn)= lim An + lim Bn then lim (Sn + 1) = s + 1
Then use an similar argument for the square root to obtain lim sn+1 = Sqrt(s + 1)
and since lim sn = lim sn+1...
Homework Statement
Let s1=1 and for n>=1 let sn+1=sqrt(sn+1)
Prove that the limit of this sequence is 1/2(1+Sqrt(5))
Homework Equations
Show that there exist an N for every \epsilon> 0, such that n>N implies
|Sn-1/2(1+Sqrt(5))|< \epsilon
The Attempt at a Solution
I can prove...