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We have that $0<b\le a$ are two positive real numbers. Their arithmetic mean is defined as $a_1=(a+b)/2$ and their geometric mean is $b_1=\sqrt{ab}$.

Firstly we have to show an that $b_1\le a_1$ iff $a=b$, so we must show both directions algebraically.

Then we define two sequences by setting $a_{n+1}=(a_n+b_n)/2$ and $b_{n+1} = \sqrt{a_nb_n}$. Using mathematical induction to show : $a\ge a_1\ge\dots \ge a_n\ge b_n\ge\dots\ge b_1\ge b$.

Prove that both series converge and show their limit.

I'm having trouble with the first step in proving the iff statement in the forward and backward direction, as well as showing the proof by induction. Also how would one determine the convergence and limit of this series?

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marked as duplicate by Arnaud D., M. Vinay, Robert Soupe, J. W. Tanner, Alex Provost Mar 14 at 19:38

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  • $\begingroup$ Do you mean series or sequence ? $\endgroup$ – Atmos Feb 27 '18 at 22:51
  • $\begingroup$ The series trivially diverge since the general term does not tend to $0$ ($a_n$ and $b_n$ are both at least $b>0$. $\endgroup$ – Bernard Feb 27 '18 at 23:00
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    $\begingroup$ Yes, i think he means sequence. $\endgroup$ – Atmos Feb 27 '18 at 23:01
  • $\begingroup$ yes, sequence, once again. $\endgroup$ – Ben French Feb 27 '18 at 23:04
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First, if $a=b$ then $$ b_1=a=a_1 $$ Now, let $a=1$ and $b=2$. Then $$ a_1=1.5 \ \text{ and }b_1=\sqrt{2} \approx 1.414 $$ Hence $a \ne b$ and $b_1 \leq a_1$. There's a problem with what you wrote isn't it ?

It might be $a_1 \leq b_1$ iff $a=b$. Indeed, suppose $$ a_1 \leq b_1 $$ Then using the fact that $x \mapsto x^2$ is increasing $$ a+b \leq 2\sqrt{ab} \Rightarrow a^2+2ab+b^2 \leq 4ab \Rightarrow \left(a-b\right)^2 \leq 0 $$ With the last inequality, we clearly see that the only possibility is $a=b$. ( which justifies directly the iff )

So in other term, we ALWAYS have $b_1 \leq a_1$.

The sequences you mention are well-known and closely linked to elliptical integrals ( discovery of Abel and Bernoulli ) because it is also linked with the length of the lemniscate. You can show that for $a=1$ and $b=\sqrt{2}$ both $(a_n)$ and $(b_n)$ converge to the same limit which is the length of a lemniscate of Bernoulli of parameter $1$ ( if i remember well ).

You can prove it by induction, but you can wisely use the property of "adjacent" sequences. https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_des_suites_adjacentes ( sorry it's french, does not know the name in english ).

You can show that $(a_n)$ increases, $(b_n)$ decreases and $a_n-b_n \underset{n \rightarrow +\infty}{\rightarrow}0$. In fact, they both converges to the same limit which is often wrote as $M\left(a,b\right)$ that stands for arithmetico-geometric mean.

You cannot find explicitly the value of the common limit $M\left(a,b\right)$. However for fun, you have beautiful properties $$ -M\left(a,b\right)=M\left(b,a\right) $$ $$ -M\left(\alpha a,\alpha b\right)=\alpha M\left(a,b\right) $$ $$ \text{min}\left(a,b\right) \leq \sqrt{ab} \leq M\left(a,b\right) \leq \frac{a+b}{2} \leq \text{max}\left(a,b\right) $$ And last but not least $$ \frac{2}{\pi}M\left(a,b\right)=\int_{0}^{\pi/2}\frac{\text{d}t}{\sqrt{x^2\cos^2(t)+y^2\cos^2(t)}} $$

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  • $\begingroup$ ok, thanks, the first part makes sense for me now, Im just trying to understand the convergence of the sequences and how they tend to zero. $\endgroup$ – Ben French Feb 27 '18 at 23:32
  • $\begingroup$ They dont tend to $0$. The difference between the two sequences tends to $0$. The two sequences both converges to the same limit, which is a real usually written as $M(a,b)$. $\endgroup$ – Atmos Feb 27 '18 at 23:34
  • $\begingroup$ ok, I was trying it out by plugging in numbers, how would you explicitly prove that both sequences converge? $\endgroup$ – Ben French Feb 27 '18 at 23:39
  • $\begingroup$ By proving what i wrote, $(b_n)$ increasing, $(a_n)$ decreasing and also $b_n-a_n \underset{n \rightarrow +\infty}{\rightarrow}0$ hence you would have the inequaltiy i wrote $a_1 \leq a_n \leq b_n \leq b_1$. And the convergence. $\endgroup$ – Atmos Feb 27 '18 at 23:47
  • $\begingroup$ ok thanks, I completed the proof using limits. $\endgroup$ – Ben French Feb 28 '18 at 0:46

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