# prove $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have same limit

I am given this problem:
let $a\ge0$,$b\ge0$, and the sequences $a_n$ and $b_n$ are defined in this way: $a_0:=a$, $b_0:=b$ and $a_{n+1}:= \sqrt{a_nb_n}$ and $b_{n+1}:=\frac{1}{2}(a_n+b_n)$ for all $n\in\Bbb{N}$

To prove is that both sequences converge and that they have the same limit. I don't know how to show this. I have spent 2 hours on this, no sign of success

• A cute question (+1) – user 1357113 Dec 30 '12 at 9:31
• @Chris'ssister, thanks :) – doniyor Dec 30 '12 at 9:33
• For a generalization to $3$ sequences (adding harmonic inequality to the mix) see this question. – Arnaud D. Nov 12 '18 at 10:39

The easy way to proceed, is to show that $b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2$ so $b_{n} \geq a_{n} \forall n \geq 2$.
Then, $a_{n+1} = \sqrt{a_n b_n} \geq a_n$ is an monotonically increasing sequence (after $n=2$).
$b_{n+1} = \frac {1}{2} (a_n + b_n) \leq b_n$ is a monotomically decreasing sequnece (after $n=2$).
Finally, $$b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2 \leq \frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} ) ( \sqrt{b_n} + \sqrt{a_n} ) = \frac {1}{2} ( b_n - a_n) \leq \frac {1}{2^n} (b_1-a_1),$$ so the difference between the sequences go to 0. Hence, these sequences converge to the same limit.

Note: Of course we could show that since $a_i \leq b_2$, the limit of $a_i$ exists (since monotonic+bounded). But I think it's more fun to jump directly to the conclusion with the final step.

• how do you know that $b_{n}$ is bounded? is it because it's a monotomically decreasing and greater than $0$ for every $n$? – Jneven Jul 13 '18 at 9:20
• and for $a_{n}$ is it formal to say that it's bounded by $b_{n}$ which is a monotomically decreasing and bounded by $0$? – Jneven Jul 13 '18 at 9:21
• @Jneven 1) I do not use that they are bounded. I have a monotonically increasing sequence $a_n$ and a monotonically decreasing sequence $b_n$ such that $b_n-a_n$ tends to 0. You should be able to conclude that these sequences converge to the same limit (Yes, I skipped a step, but this should be obvious). – Calvin Lin Jul 27 '18 at 15:47
• 2) (I'm not certain if you're stating all of the information of the sequence, or relying on the problem) If that's all you have, then no. E.g. The sequnce $0, 1, 0, 1, 0, 1, ...$ is bounded the monotonically decreasing sequence $2,2,2,2, ...$ and by 0, but doesn't converge. In this case,we have $a_n$ is monotonic, so yes it convergs. – Calvin Lin Jul 27 '18 at 15:47
• For the record, someone asked a question on this answer : math.stackexchange.com/questions/927192/… – Arnaud D. Mar 14 at 13:10

This is not an answer. Just a long comment which gives a classical formula of Gauss for this common limit.

Let $a$, $b$ be real numbers with $a<b$. We define their arithmetic-geometric mean $M(a,b)$ using the sequences $a_n$, $b_n$ defined as follows: $a_0=a$, $b_0=b$ and inductively

$$a_{n+1}={ a_n +b_n \over 2}, \qquad b_{n+1}=\sqrt {a_n b_n }\ .$$

The two sequences converge to a unigue common limit $M(a,b)$. Gauss discovered a beautiful formula for $M(a,b)$ which can be expressed as an elliptic integral

$${ 1 \over M(a,b)} = { 2 \over \pi} \int_0^{ { \pi \over 2}} { d \theta \over \sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta}} \ .$$

• assuming the use of integrals isn't relevant for me (this question was introduced to me in the class of "calculus 1" where integrals are not in the syllabus) how would you bound $b_{n}$? – Jneven Jul 13 '18 at 9:24
• Another formula can be found here : math.stackexchange.com/questions/2301340/… – Arnaud D. Mar 14 at 13:05

If $a_n\to l$ and $b_n \to s$ then from $b_{n+1}=\frac{1}{2}(a_n+b_n)$ we get $s=\frac{1}{2}(l+s)\cdots$
To prove that they converge use arithmetic and geometric mean inequality to show (that $a_n\leq b_n$ so ...) they are monotonic ($a_n$ is increasing and $b_n$ decreasing). From $a_n\leq b_n$ conclude they are bounded.

• Pambos, thanks, now i got it – doniyor Dec 30 '12 at 7:28

If $I(a, b) =\int_0^{\pi/2} \dfrac{dt}{\sqrt{a^2\cos^2t+b^2\sin^2t}}$, using Landen's transformation (https://en.wikipedia.org/wiki/Landen%27s_transformation) you can show that $I(a, b) =I(\dfrac{a+b}{2}, \sqrt{ab})$.
Using the well-known inequalities in Calvin Lin's answer, this shows that $I(a, b) =I(AGM(a, b), AGM(a, b)) =\dfrac{\pi}{2AGM(a, b)}$, which gives a quickly converging way of computing the complete elliptic integral of the first kind.