Induction on an inequality involving sequences

I just can't get the following question. Can someone help me out?In summary, the conversation discusses a question involving sequences and their convergence. The question is broken down into three parts, with the first part requiring a proof that the sequences are bounded and increasing or decreasing. The second part uses the monotonic sequence theorem to prove the convergence of both sequences, and the third part proves that the two sequences have the same limit, which is referred to as the arithmetic-geometric mean of the initial values. The conversation also includes some attempts at solving the question, including using specific cases and induction.
  • #1
Benny
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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.
 
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  • #2
Here are a few comments. If 0< a< b then
1) 0< a2< ab so 0< a< [tex]\sqrt{ab}[/tex]
That gives you "a2< a1".
2) 0< a+ b< 2b so 0< (a+b)/2< b.
That gives you "b2< b1".
Now, assume that [tex]a_n \le a_{n + 1} \le b_{n + 1} \le b_n [/tex]
for some k and use the same calculations as above to show that
[tex]a_{n+1} \le a_{n + 2} \le b_{n + 2} \le b_{n+1} [/tex]
 
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  • #3
Thanks for the help HallsofIvy.
 

1. What is induction on an inequality involving sequences?

Induction on an inequality involving sequences is a mathematical technique used to prove that a given inequality holds for all values of n, where n is a natural number. It involves using the principle of mathematical induction to prove the inequality for the base case and then showing that if the inequality holds for some arbitrary value of n, it also holds for n+1.

2. Why is induction on an inequality involving sequences useful?

This technique is useful because it allows us to prove that an inequality holds for infinitely many values of n, without having to check each value individually. It also provides a systematic approach to proving inequalities involving sequences.

3. What is the principle of mathematical induction?

The principle of mathematical induction states that if we can prove that a statement holds for a base case (usually n=1 or n=0) and then show that if it holds for some value of n, it also holds for n+1, then the statement must hold for all natural numbers.

4. How do you use induction on an inequality involving sequences?

To use induction on an inequality involving sequences, we first prove the base case (n=1 or n=0) and then assume that the inequality holds for some arbitrary value of n. We then use this assumption to show that the inequality also holds for n+1. This completes the inductive step and proves that the inequality holds for all natural numbers.

5. Can induction be used to prove all inequalities involving sequences?

No, induction can only be used to prove inequalities that involve natural numbers. It cannot be used to prove inequalities involving real numbers or other mathematical objects. In some cases, other techniques such as proof by contradiction or direct proof may be more suitable for proving inequalities.

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