Arithmetic-Geometric limit $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n$

For $$x,y>0$$, define two sequences $$(x_n)$$ and $$(y_n)$$ by $$x_1=x,y_1=y$$ and $$x_{n+1}=(x_n+y_n)/2$$ and $$y_{n+1}=\sqrt{x_ny_n}$$. Prove that $$\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n= \dfrac{\pi}{\int_0^\pi \dfrac{d\theta}{\sqrt{x^2 \cos^2\theta + y^2\sin^2\theta}}}.$$

I think it might be easier to prove $$\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n.$$ Let the LHS of this equation be denoted $$L$$ and let the RHS be denoted $$M$$. By the AM-GM inequality and induction, $$x_{n}\leq y_n$$ for all $$n\ge 2$$. It could be useful to define a new sequence with a limit that's easier to evaluate. We have $$L\leq M$$ by limit properties, so we just need to show $$L\ge M$$ to get $$L=M$$. Suppose for a contradiction that $$L < M.$$ Then by definition, there exists $$N$$ so that for all $$n\ge N, x_n < \frac{L+M}2$$ and $$y_n > \dfrac{L+M}2.$$ How can I proceed from here?

As for showing it equals an expression involving a given integral, I think the integral is actually fairly hard to compute explicitly, so one should use some properties of the sequences to show the desired equality.

• You don't actually need a contradiction. Observe that: $$L = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \frac{x_n+y_n}{2} = \frac{L+M}{2}$$ This implies that $L = M$. As for the integral, I feel like doing some trigonometric substitution within the sequences themselves might get you somewhere Commented Oct 2, 2022 at 2:16
• For the integral formula, proceed via the proof due to Gauss. Start from the integral and devise a substitution that will relate the integral $I(x,y)$ to a corresponding integral $I(\frac{1}{2}(x+y),\sqrt{xy})$.
– Jam
Commented Oct 2, 2022 at 2:27
• The case $x=y$ is trivial. If $x\ne y$ then WLOG $x>y.$ Since $u>v>0\implies u>(u+v)/2>\sqrt {uv}>v,$ we have $x_n>x_{n+1}>y_{n+1}\ge y_1$ (the last inequality by induction on $n$) .So $x_n$ decreases & is bounded below by $y_1$. So $x_n$ converges to a limit. The limit is called The Arithmetic-Geometric Mean, which is also the title of a book about it, with some very deep results. BTW the convergence is extremely fast. Commented Oct 2, 2022 at 5:02
• @AbhijeetVats Doesn't your statement assume existence. I agree that IF the limits exist, they will trivially be equal, but their existence is non-trivial. Commented Oct 2, 2022 at 19:34
• The integral at the end is tricky. See paramanands.blogspot.com/2009/08/… Commented Oct 7, 2022 at 4:04

I will handle the integral at the end using the transformation given by Gauss. An alternative transformation is available on my blog (linked in comments to question).

Let us write $$I(x, y) =\int_0^{\pi/2}\frac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}\tag{1}$$ for $$x, y>0$$ and we prove $$I(x, y) =I\left(\frac{x+y} {2},\sqrt {xy} \right) \tag{2}$$ Gauss used the substitution $$\sin t=\frac{2x\sin u} {x+y+(x-y) \sin^2u}\tag{3}$$ to establish $$(2)$$. The substitution maps the interval $$[0,\pi/2]$$ to $$[0,\pi/2]$$.

Let us first observe that $$\cos t=\frac{\sqrt{(x+y)^2+(x-y)^2\sin^4u-2(x^2+y^2)\sin^2u}} {x+y+(x-y) \sin^2u}$$ which can be rewritten as $$\frac{\sqrt{4a^2+4(a^2-b^2)\sin^4u-4(2a^2-b^2)\sin^2u}} {x+y+(x-y) \sin^2u}$$ where $$2a=x+y,b^2=xy$$. The above can be further simplified as $$\cos t=\frac{2\cos u\sqrt{a^2\cos^2u+b^2\sin^2u}}{x+y+(x-y)\sin^2u}\tag{4}$$ Next we have $$x^2\cos^2t+y^2\sin^2t=\frac{4x^2\cos^2u(a^2\cos^2u+b^2\sin^2u)+4x^2y^2\sin^2u}{(x+y+(x-y)\sin^2u)^2}$$ which can be rewritten as $$\frac{x^2[(x+y)^2\cos^4u+4xy\sin^2u\cos^2u+4y^2\sin^2u]}{(x+y+(x-y)\sin^2u)^2}$$ Replacing $$\cos^2u$$ with $$1-\sin^2u$$ we get $$\sqrt{x^2\cos^2t+y^2\sin^2t}=\frac{x(x+y-(x-y)\sin^2u)}{x+y+(x-y)\sin^2u}\tag{5}$$ Differentiating equation $$(3)$$ with respect to $$u$$ we get $$\cos t\cdot \frac{dt}{du} =\frac{2x\cos u(x+y-(x-y) \sin^2u)} {(x+y+(x-y)\sin^2u)^2} \tag{6}$$ Using $$(4),(5),(6)$$ we get $$\int_{0}^{\pi/2}\frac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}=\int_{0}^{\pi/2}\frac{du}{\sqrt{a^2\cos^2u+b^2\sin^2u}}$$ or $$I(x,y) =I(a,b)$$.

Thus we get $$I(x, y) =I(x_n, y_n)$$ and taking limits as $$n\to\infty$$ we get $$I(x, y) =\lim_{n\to\infty} I(x_n, y_n) =I(\lim_{n\to\infty} x_n, \lim_{n\to\infty} y_n) =I(L, L)$$ We have used the fact that $$I(x, y)$$ is a continuous function of $$x, y$$ and that the sequences $$x_n, y_n$$ tend to a common limit, say $$L$$.

Now $$I(L, L) =\pi/(2L)$$ it follows that $$L=\frac{\pi} {2I(x,y)}=\dfrac{\pi}{\displaystyle \int_0^{\pi}\dfrac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}}$$

• Thanks for taking the time to write out this answer even though you already made a blog post on an alternative approach. Commented Oct 8, 2022 at 19:03

Putting together (essentially) the proof from comments that the limits exist and are equal, by @AbhijeetVats and @DanielWainfleet:

If $$x=y$$, then $$x_n=y_n=x=y$$ for all $$n$$, so obviously the limits exist and are equal.

If $$x \neq y$$, then $$y_2 < x_2$$ by the AM-GM inequality. Also note $$y_2 > 0$$. Now prove by induction that for every $$n \geq 2$$,

$$y_2 \leq y_n < y_{n+1} < x_{n+1} < x_n \leq x_2$$

Since $$x_n > y_n > 0$$, $$y_{n+1} = \sqrt{x_n y_n} > \sqrt{y_n y_n} = y_n$$. Since $$y_n \geq y_2$$, this also shows $$y_{n+1} > y_2$$.

Since $$y_n < x_n$$, $$x_{n+1} = \frac{x_n+y_n}{2} < \frac{x_n+x_n}{2} = x_n$$.

And by the AM-GM inequality again, $$y_{n+1} < x_{n+1}$$. The induction proof is complete.

So $$(x_n)$$ is a strictly decreasing sequence bounded below by $$y_2$$, and therefore converges to a real value $$L = \lim_{n \to \infty} x_n$$. And $$(y_n)$$ is a strictly increasing sequence bounded above by $$x_2$$, and therefore converges to a real value $$M = \lim_{n \to \infty} y_n$$.

But then

$$L = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \frac{x_n+y_n}{2} = \frac{M+L}{2}$$

which implies that $$L=M$$:

$$\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n$$

Just showing that $$x_1=x, y_1=y, x_{n+1}=\frac{x_n+y_n}{2}, y_{n+1}=\sqrt{x_ny_n}.$$ Show that $$\displaystyle \lim_{n\ \to \infty} x_n = \lim_{n \to \infty} y_n.$$

let $$x_n-y_n=d_n.$$

Then, $$d_n\geq 0$$, because of AM-GM inequality.

$$x_{n+1}=\frac{x_n+y_n}{2} \leq \frac{x_n+x_n}{2}=x_n.$$

$$y_{n+1}=\sqrt{x_ny_n} \geq \sqrt{y_ny_n}=y_n.$$

$$\therefore x_n-y_n \geq x_{n+1}-y_{n+1}.$$

If $$x_{n+1}=x_n$$ and $$y_{n+1}=y_n$$, $$x_n=y_n.$$

That means that until $$x_n-y_n \neq 0$$, $$x_n-y_n > x_{n+1}-y_{n+1}.$$

Done. We got that $$d_n$$ is a bounded and decreasing function.

Therefore, we get $$\displaystyle \lim_{n \to \infty} d_n=0.$$

• A couple questions raised by your argument: (1) Does $\lim d_n = 0$ imply $\lim x_n=\lim y_n$? (2) Does the fact $d_n$ is strictly decreasing and positive for $n\geq 2$ imply $\lim d_n=0$? Commented Oct 4, 2022 at 4:56
• Adding to @BrianMoehring's point (1): If $\lim x_n$ and $\lim y_n$ exist, then $\lim d_n=0$ implies the limits are equal. But have you shown those limits do exist? Commented Oct 4, 2022 at 13:08

This limit is known as the arithmetic-geometric mean of $$x$$ and $$y$$, denoted by $$AGM(x,y)$$ and goes back to Lagrange. It is appears in the estimation of the period of periodic solutions of the pendulum equation $$\ddot{x}+ k\sin x=0$$ where elliptic integrals such as the one given in the OP appears.

Here is a summary of the convergence of the sequences in the OP and also the speed of converge.

• If $$0\leq y_0=y_0$$ then convergence is trivial and as the sequence is stationary: $$x_n=x_0=y_n=x_0$$

• If $$y_0=0$$ and $$x_0>0$$, then clearly $$y_n=0$$ for all $$n$$ and $$x_n=2^{-n}x_0\xrightarrow{n\rightarrow\infty}0$$.

• Without loss of generality, assume that $$0, for if $$0 and $$x_0\neq y_0$$ then $$y_1=\sqrt{x_0y_0}<\frac{x_0+y_0}{2}=x_1$$ By induction, given that $$y_{n-1}, we have that \begin{align} y_n&=\sqrt{x_{n-1}y_{n-1}}>y_{n-1}\\ x_n&=\frac{x_{n-1}+y_{n-1}}{2} Thus for some $$y_0, $$y_n\nearrow y$$ and $$x_n\searrow x$$ and so, $$x=\frac{x+y}{2}$$ whence $$x=y$$. Finally, notice that \begin{align} x_n-y_n=\frac{x_{n-1}+y_{n-1}}{2}-\sqrt{x_{n-1}y_{n-1}}<\frac{x_{n-1}+y_{n-1}}{2}-y_n=\frac{x_{n-1}-y_{n-1}}{2} \end{align} whence it follows that \begin{align} AMG(x_0,y_0)-y_n&<\frac{x_0-y_0}{2^n}\\ x_n-AMG(x_0,y_0)&<\frac{x_0-y_0}{2^n} \end{align} This shows that the convergence to the limit $$AMG(x_0,y_0)$$ is actually quite fast. It is readily seen that \begin{align} AMG(rx,ry)&=r\,AMG(x,y), \qquad x,y,r\geq0\\ AMG(x,y)&= AMG\big(\frac{x+y}{2},\sqrt{xy}\big) \end{align}

The integral expression of the limit $$AMG(x_0,y_0)$$ was obtained by Gauss (see this Wikipedia [article]((https://en.wikipedia.org/wiki/Arithmetic–geometric_mean) on his treatment of the Cayley elliptic integral $$I(a, b):=\int^{\pi/2}_0\frac{d\theta}{\sqrt{a^2 \cos^2\theta + b^2\sin^2\theta}}$$ Gauss introduces the change of variables \begin{align} \sin \theta ={\frac {2a\sin \theta'}{(a+b)+(a-b)\sin ^{2}\theta'}} \end{align} and obtains that $$I(a, b)=I\big(\frac{a+b}{2},\sqrt{ab}\big)$$ $$I$$ is a continuous function on $$(0,\infty)\times(0,\infty)$$ and so $$I(a, b)=I(a_n,b_n)$$ where $$a_n=\frac{a_{n-1}+ b_{n-1}}{2}$$ and $$b_n=\sqrt{a_{n-1}b_{n-1}}$$ and so, $$I(a, b)=\lim_n I(a_n, b_n)=I(AMG(a, b), AMG(a, b))=\frac{\pi}{2}AMG(a, b)$$