Let s1 = √(6), s2 = √(6+√(6)), s3 = √(6+√(6+√(6))), and in general define sn+1 = √(6+sn). Prove that (sn) converges, and find its limit.
*note: "√" means "the square root of"
show that the sequence is monotonic (increasing) and bounded. this will show that it has a limit (by what theorem?) you can prove both pretty easily by induction. do you see what to prove for each of these?
to find the limit, note that , call this limit .
then, since we have
you want the positive solution (why?)
I know that I have to prove that sn+1 > sn, which will prove that it is monotonically increasing. Then I would have to find an upper bound, U, for it and prove that sn < U and sn+1 < U. This would prove that it is bounded. Then, by the monotone convergence theorem, the sequence must be convergent and have a limit.
I am having trouble putting this all together in a proper proof.
O.K.
Lets take the difference .................................................. ......................................1
Now if you multiply (1) by ,and do couple of calculations you get the formula:
.................................................. .....................................2
In this formula all the terms are +ve except the factor .
But since we can prove that for all n, then by using (2) we get .
Hence the sequence is increasing and bounded from above by 3.
So it is left to you to prove for all ,n and for all ,n.
Up to now we have proved :
a) IF the sequence converges ,then it must converge to 3 (because for all ,n)
b) The sequence converges.
BY (a) and (b) and using M.Ponens ,we have:
The limit of the sequence is 3.
Note usually the finding of a limit for this type of sequences consists of those two parts.
In the first part WE Assume that the sequence converges,and find its limit .
In the 2nd part we prove convergence