On sequences of means

[gugo82]

This is an elementary exercise on recursive sequences.

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All the stuff to be useful in finding the solution is written here.

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Exercise:

Let [tex]$x,y\in [0,+\infty[$[/tex], [tex]$p>q \in [0,+\infty[$[/tex] and set:

[tex]$\begin{cases} a_1:=M_p(x,y)\\ b_1:=M_q(x,y) \\ a_{n+1}:=M_p(a_n,b_n) &\text{, for each } n\in \mathbb{N}\\ b_{n+1}:=M_q(a_n,b_n) &\text{, for each } n\in \mathbb{N}\end{cases}$[/tex].

Prove that both sequences [tex]$(a_n)$[/tex] and [tex]$(b_n)$[/tex] converge and that they have the same limit.

[Leonardo891]

Thank you Gugo for your exercises and to keep alive this section of the forum!

[gugo82]

Ah, OK.
I didn't understand for what part of the proof you were using induction; now it is clear that you used it while looking for upper and lower bounds.
The proof is correct.

Thank you Leonardo.


P.S.: I really wasn't expecting an answer so soon... Your proof was a nice surprise!

[Leonardo891]

Thank you for correcting my mistakes in English!
You're right: I should be more thorough.
Then

The sequences $a_n$ and $b_n$ are bounded.
I know that $a_1=M_p(x,y),b_1=M_q(x,y) \in ] x,y [ $
By inductive hypothesis $a_n,b_n \in ]x,y[ $ and, since $a_{n+1}=M_p(a_n,b_n),b_{n+1}=M_q(a_n,b_n) \in ]text{min}(a_n,b_n),text{max}(a_n,b_n)[$ we have $a_{n+1},b_{n+1} \in ]x,y[$.

Now from pMqM I know $AA n \in NN^+, b_nq).

The sequence $a_n$ is strictly decreasing.
$AA n \in NN^+, a_{n+1}=M_p(a_n,b_n) \in ]b_n,a_n[ => a_{n+1}
The sequence $a_n$ is strictly increasing.
$AA n \in NN^+, b_{n+1}=M_q(a_n,b_n) \in ]b_n,a_n[ => b_{n+1}>b_n$

Now the rest of my previous try.
Now I can say that $a_n$ and $b_n$ are convergent and we have $l' :=lim_{n-> \infty} a_n = text{inf} {a_n, n \in NN^+ }$ and $l'':=lim_{n-> \infty} b_n = text{sup} {b_n, n \in NN^+ }$.
Since $AA n \in NN^+, b_n Let's suppose, for absurd, that l'' Since $a_{n+1}=M_p(a_n,b_n)$ passing to the limit, for the continuity of the mean, we have $l'=M_p(l',l'')$ than $l'' Than we have l'=l''.

Now the proof should be ok, at least I hope!

[gugo82]

@Leonardo: "Then" (allora) is waaay more appropriate than "than" (che, quanto).

A direct proof of the equality [tex]$l^\prime =l^{\prime \prime}$[/tex] is the following:

[tex]$b_{n+1}=M_q(a_n,b_n) \quad \Rightarrow \quad a_n = \left\{ 2\ b_{n+1}^q -b_n^q\right\}^{\frac{1}{q}} \quad \Rightarrow \quad l^\prime =\lim a_n =\lim \left\{ 2\ b_{n+1}^q -b_n^q \right\}^{\frac{1}{q}} =l^{\prime \prime}$[/tex].

But I can't catch why you wrote:

"Leonardo89":Then we have $x
I mean, the inequality in the middle of [tex]$x < b_n < b_{n+1} < a_{n+1} < a_n < y$[/tex]: how can you prove it from the inductive hypothesis [tex]$b_n < a_n$[/tex]?
Actually the inequality [tex]$b_n \leq a_n$[/tex] follows directly from (pMqM) in here; so maybe you don't want to prove this fact by induction...
Please, can you formalize in a better way your proof? Thanks in advance.

[Leonardo891]

Here's my try.
If $x=y$ than $AA n \in NN^+, a_n=b_n=x=y$.
Than I can assume, without loss of generality, that x I will show that both $a_n$ and $b_n$ are bounded, that $a_n$ is strictly decreasing and that $b_n$ is strictly increasing.
It's clear that $AA r,s \in [0,+ \infty[, r r Than we have $x Now I can say that $a_n$ and $b_n$ are convergent and we have $l' :=lim_{n-> \infty} a_n = text{inf} {a_n, n \in NN^+ }$ and $l'':=lim_{n-> \infty} b_n = text{sup} {b_n, n \in NN^+ }$.
Let's suppose, for absurd, that l'' Since $a_{n+1}=M_p(a_n,b_n)$ passing to the limit, for the continuity of the mean, we have $l'=M_p(l',l'')$ than $l'' Than we have l'=l''.
I hope that my English isn't too much bad.