Bourbaki

Exercise 16, § 5

General Topology, IV, 392

Let x0,y0x_0, y_0 be two real numbers such that 0<y0≤x00 < y_0 \leq x_0. For n>0n > 0 we define two sequences (xn),(yn)(x_n), (y_n) inductively by the relations

xn+1=(xn+yn)/2,yn+1=xnyn. x_{n+1} = (x_n + y_n)/2, \quad y_{n+1} = \sqrt{x_n y_n}.

Show that the two sequences (xn),(yn)(x_n), (y_n) tend to the same limit α\alpha (the "arithmetico-geometric mean" of x0,y0x_0, y_0); also there exist numbers a>0a > 0 and γ\gamma such that 0<γ<10 < \gamma < 1 and xn−yn≤aγ2nx_n - y_n \leq a \gamma^{2^n} for each nn [observe that xn+1−yn+1=(xn−yn)2/4(xn+1+yn+1)x_{n+1} - y_{n+1} = (x_n - y_n)^2 / 4(x_{n+1} + y_{n+1})].

Solution

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Transcribed from the printed volume. Not translated.