# How to prove two sequence have a common limit.

We have \begin{align} U_{0} &= 1 &&\text{and} & V_{0} &= 2 \\ U_{n+1} &= \frac{U_{n}+V_{n}}{2} &&\text{and} & V_{n+1} &= \sqrt{U_{n+1}V_n} \end{align} How to prove two sequence have a common limit?

I found $(U_n)$ is increasing and $(V_n)$ is decreasing but I don't know how to do with $\lim (V_n) - (U_n)$

• Do you want $U_{n+1} = \frac{U_n+V_n}2$ instead? – Somos Sep 15 '18 at 20:55
• @Somos I thought I should use only that limit $V_n - U_n$ but yeah if it's possible to resolve with it , how it works ? – KEVIN DLL Sep 15 '18 at 20:58
• I ask because you never tell us what $\,b\,$ is supposed it be. Is it a real number? – Somos Sep 15 '18 at 21:00
• @Somos yes i just edit it , – KEVIN DLL Sep 15 '18 at 21:02
• What is an "adjacent" sequence? Do you mean convergent? – Somos Sep 15 '18 at 21:09

Since $$U_n$$ is increasing and $$V_n$$ is decreasing, all you have to show is that the difference $$V_n-U_n$$ has limit $$0$$. But, the limit exists since both sequences are monotone and bounded, and rewriting the equation $$U_{n+1}=(U_n+V_n)/2$$ to $$\,V_n = 2 U_{n+1}-U_n \,$$ shows the two limits must be equal.

Given a diameter one circle, its circumference is $$\,\pi.\,$$ Archimedes calculated the perimeters of inscribed and circumscribed regular polygons to find upper and lower bounds for $$\,\pi.\,$$ Let $$\,a(n)\,$$ be the perimeter of the circumscribed regular polygon of $$\,n\,$$ sides, and $$\,b(n)\,$$ the perimeter of the inscribed regular polygon of $$\,n\,$$ sides. We have that $$\,a(2n)\,$$ is the harmonic mean of $$\,a(n)\,$$ and $$\,b(n)\,$$ and $$\,b(2n)\,$$ is the geometric mean of $$\,a(2n)\,$$ and $$\,b(n).\,$$

We can start with triangles where $$\,n=3\,$$ and find $$\,a(3) = 3\sqrt{3}\,$$ and $$\,b(3) = a(3)/2.\,$$ We then keep doubling the number of sides indefinitely. The connection between the two recursions is that $$\, U_n = 3\sqrt{3}/a(3\,2^n)\,$$ and $$\, V_n = 3\sqrt{3}/b(3\,2^n)\,$$ since the recursions and initial values for $$\, U_n, V_n \,$$ come from those for $$\,a(n), b(n).\,$$ The common limit of $$\,a(n), b(n)\,$$ is $$\,\pi,\,$$ thus the common limit of $$\, U_n, V_n\,$$ is $$\,3\sqrt{3}/\pi.\,$$

If we start with squares where $$\,n=4\,$$ we find that $$\, a(4) = 4\,$$ and $$\,b(4) = 2\sqrt{2}.\,$$ Now let $$\, U_n := 4/a(4\, 2^n)\,$$ and $$\, V_n := 4/b(4\, 2^n).\,$$ Then $$\, U_0 = 1, \,\, V_0 = \sqrt{2} \,$$ and the common limit is $$\, 4/\pi.\,$$ The general values are $$\, b(n) = n\,\sin(\pi/n),\,$$ and $$\, a(n) = n\,\tan(\pi/n).$$

Given that $$u_{n+1}=\frac{u_n+v_n}2\quad\text{and}\quad v_{n+1}=\sqrt{\vphantom{1}u_{n+1}v_n}\tag1$$ it is easy to derive that $$u_{n+1}^2-v_{n+1}^2=\frac{u_n^2-v_n^2}4\tag2$$ Thus, induction says that $$u_n^2-v_n^2=\frac{u_0^2-v_0^2}{4^n}\tag3$$ Furthermore, $$u_{n+1}-u_n=\frac{v_n-u_n}2\quad\text{and}\quad v_{n+1}^2-v_n^2=\frac{u_n-v_n}2v_n\tag4$$ Case $\boldsymbol{u_0\gt v_0}$

Equation $(3)$ says $u_n\gt v_n$, and $(4)$ says that $u_n$ is decreasing and $v_n$ is increasing. Since, $u_n\gt v_n\gt v_0$, $u_n$ is decreasing and bounded below. Since $v_n\lt u_n\lt u_0$, $v_n$ is increasing and bounded above.

Case $\boldsymbol{u_0\lt v_0}$

Equation $(3)$ says $u_n\lt v_n$, and $(4)$ says that $u_n$ is increasing and $v_n$ is decreasing. Since, $u_n\lt v_n\lt v_0$, $u_n$ is increasing and bounded above. Since $v_n\gt u_n\gt u_0$, $v_n$ is decreasing and bounded below.

In either case, both $u_n$ and $v_n$ converge. $(3)$ says that their limits are the same.

If $U_n,V_n\in[1,2]$,

$$U_n+V_n\in[2,4]\implies U_{n+1}\in[1,2]$$

and

$$U_{n+1}V_n\in[1,4]\implies V_n\in[1,2].$$

Hence both sequences are bounded and both converge.

And if the limits exist, they are equal. Indeed,

$$U=\frac{U+V}2,V=\sqrt{UV}$$ imply $U=V$.

• Do you not need some monotony argument? Maybe I overlooked something, but, wouldn't it be also possible that the series just circle around some values? – Imago Sep 17 '18 at 16:16
• @Imago: re-read the post, monotonicity is already granted. – Yves Daoust Sep 17 '18 at 16:22
• I take this as a yes. It's needed. – Imago Sep 17 '18 at 16:24