The sequence is: a(n+1)=4+2(a(n))^(1/3) with a(1)=1
Computing this clearly shows the sequence converges to 8, but how do I prove this?
In earlier parts I have already proved the sequence is bounded, and increasing, so that proves that the sequence does converge, but now I need to prove what the limit is.
Thanks
To prove that it converges to 8, you will first need to prove it converges. After that, you can use Plato's idea. But before you can call the limit "L", you must show that the limit exists.
For this problem and "iterative" problems in general, it is typically easiest to use the "monotone convergence theorem": "If a sequence of real number is increasing and has an upper bound, it converges".
So we need to prove two things: that this sequence is has an upper bound and that it is increasing. Since we want this (increasing) sequence to converge to 8, a good candidate for an upper bound is 8.
First, it is clear that all [itex]a_n[/itex] are positive.
Prove, for all n, .
And, again because this is "iterative", it is easiest to use 'proof by induction'. When i= 1, so the base case is true. Assume for some k and look at . . Because , . I'll let you finish this.
Prove, for all n, .
We have and [tex]a_2= 4+ 2(1^{1/2})= 4+ 2= 6> 1[/itex] so the base case is true.
Assume that for some k. Then . Because [itex]0< a_k< a_{k+1}[/itex], [itex]0< a_k^{1/3}< a_{k+1}^{1/3}[/itex]. Again, it should be easy to finish this.
The recursive relation can be written as...
(1)
The function is illustrated here...
There is only one 'attractive fixed point' in and, because is...
(2)
... any will produce a sequence that monotonically converges at . The 'rigorous' demonstration of that is obtained considering that the sequence convergences at if and only if converges the series...
(3)
Now it is easy to see that if (2) is satisfied, then the ratio test extablishes that (3) converges...
Kind regards