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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 $\lim (V_n) - (U_n)$

Can someone help me, please?

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  • $\begingroup$ Do you want $U_{n+1} = \frac{U_n+V_n}2$ instead? $\endgroup$ – Somos Sep 15 '18 at 20:55
  • $\begingroup$ @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 ? $\endgroup$ – KEVIN DLL Sep 15 '18 at 20:58
  • $\begingroup$ I ask because you never tell us what $\,b\,$ is supposed it be. Is it a real number? $\endgroup$ – Somos Sep 15 '18 at 21:00
  • $\begingroup$ @Somos yes i just edit it , $\endgroup$ – KEVIN DLL Sep 15 '18 at 21:02
  • $\begingroup$ What is an "adjacent" sequence? Do you mean convergent? $\endgroup$ – Somos Sep 15 '18 at 21:09
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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 $\, a(n) = n\,\tan(\pi/n).$

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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.

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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 $U=V$.

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  • $\begingroup$ 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? $\endgroup$ – Imago Sep 17 '18 at 16:16
  • $\begingroup$ @Imago: re-read the post, monotonicity is already granted. $\endgroup$ – Yves Daoust Sep 17 '18 at 16:22
  • $\begingroup$ I take this as a yes. It's needed. $\endgroup$ – Imago Sep 17 '18 at 16:24

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