- #1
Benny
- 584
- 0
I just can't get the following question. Can someone help me out?
Q. Let a < a_1 < b_1 and define [tex]a_{n + 1} = \sqrt {a_n b_n } ,b_{n + 1} = \frac{{a_n + b_n }}{2}[/tex] .
a) Prove that [tex]a_n \le a_{n + 1} \le b_{n + 1} \le b_n [/tex] for all n.
b) Deduce that the sequences {a_n} and {b_n} both converge.
c) Prove that they have the same limit. (This limit is called the arithmetic-geometric mean of a_1 and b_1)
Here are my attempts
b) I'll do this one first. By the result of part 'a' {a_n} is increasing and bounded above by b_n while {b_n} is a decreasing sequence and bounded below by a_n. So by the monotonic sequence theorem both sequences converge. Not sure about the argument there, I might have missed a few important points.
a) There might ways other than induction to do this but I can't think of any. So I need to start off with a specific case.
For n = 1: I can get a_1 <= a_(2) but not the b_(2) <= b_(1) part 'properly' and I can't get the a_2 <= b_2 bit at all.
Q. Let a < a_1 < b_1 and define [tex]a_{n + 1} = \sqrt {a_n b_n } ,b_{n + 1} = \frac{{a_n + b_n }}{2}[/tex] .
a) Prove that [tex]a_n \le a_{n + 1} \le b_{n + 1} \le b_n [/tex] for all n.
b) Deduce that the sequences {a_n} and {b_n} both converge.
c) Prove that they have the same limit. (This limit is called the arithmetic-geometric mean of a_1 and b_1)
Here are my attempts
b) I'll do this one first. By the result of part 'a' {a_n} is increasing and bounded above by b_n while {b_n} is a decreasing sequence and bounded below by a_n. So by the monotonic sequence theorem both sequences converge. Not sure about the argument there, I might have missed a few important points.
a) There might ways other than induction to do this but I can't think of any. So I need to start off with a specific case.
For n = 1: I can get a_1 <= a_(2) but not the b_(2) <= b_(1) part 'properly' and I can't get the a_2 <= b_2 bit at all.