Question about prove when {xn}\{x_n\} and {yn}\{y_n\} convergent limxn=limyn\lim x_n = \lim y_n where xn+1=xn+yn2x_{n+1}=\frac{x_n+y_n}2, yn+1=2xnynxn+yny_{n+1}=\frac{2x_ny_n}{x_n+y_n}

Problem 4\mathbf{4} If xn+1=xn+yn2x_{n+1}=\frac{x_n+y_n}2, yn+1=2xnynxn+yny_{n+1}=\frac{2x_ny_n}{x_n+y_n} for n=1,2,…,nn=1,2,\ldots,n, prove that both {xn}\{x_n\} and {yn}\{y_n\} are convergent, and limn→∞xn=limn→∞yn\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n.

Solution: Note that 4xnyn≤(xn+yn)24x_ny_n\le(x_n+y_n)^2 Hence, yn+1xn+1=4xnyn(xn+yn)2≤1\frac{y_{n+1}}{x_{n+1}}=\frac{4x_ny_n}{(x_n+y_n)^2}\le 1 That is, xn≤yn\color{crimson}{x_n\le y_n}. Also, xn+1−xn=xn+yn2−xn=yn−xn2≥0x_{n+1}-x_n=\frac{x_n+y_n}2-x_n=\frac{y_n-x_n}2\ge 0 implying that {xn}\{x_n\} is monotonically increasing. Similarly, yn+1yn=2xnxn+yn≤2xn2xn=1\frac{y_{n+1}}{y_n}=\frac{2x_n}{x_n+y_n}\le\frac{2x_n}{2x_n}=1 implying that {yn}\{y_n\} is monotonically decreasing. Now we have x1≤x2≤⋅≤xn≤⋅≤yn≤⋯≤y2≤y1x_1\le x_2\le\cdot\le x_n\le\cdot\le y_n\le\cdots\le y_2\le y_1 By the monotone convergence theorem, it follows immediately that both {xn}\{x_n\} and {yn}\{y_n\} are convergent. Let limn→∞xn=x,limn→∞yn=y\lim_{n\to\infty}x_n=x,\lim_{n\to\infty}y_n=y Taking limits on both sides of xn+1x_{n+1} will yield that x=x+y2x=\frac{x+y}2 which implies that limn→∞xn=x=y=limn→∞yn\lim\limits_{n\to\infty}x_n=x=y=\lim\limits_{n\to\infty}y_n

(Original image here.)

I have a question about how I can know that xn≤ynx_n\le y_n (in red above).

I know that yn+10x_{n+1}>0, and at other points in the argument there appears to be an implicit assumption that xnx_n and yny_n are positive. This will be the case if we add the assumption that x1≥0x_1\ge 0 and y1>0y_1>0. A similar argument works if xnx_n and yny_n are always both negative.

The second step should then be

xn+1−xn=xn+yn2−xn=yn−xn2≤0,x_{n+1}-x_n=\frac{x_n+y_n}2-x_n=\frac{y_n-x_n}2\le 0\;,

so that {xn}\{x_n\} is monotonically decreasing. The next step then becomes


so that {yn}\{y_n\} is monotonically increasing, and we have

y1≤y2≤…yn≤…≤xn≤…x2≤x1.y_1\le y_2\le\ldots y_n\le\ldots\le x_n\le\ldots x_2\le x_1\;.

It still follows from the monotone convergence theorem that both sequences are convergent, and we can let their limits be xx and yy, as in the given solution. The final step is also correct, but I’ll expand on it to make clearer just what is being done. The point is that


so that


Thus, 2x=x+y2x=x+y, and x=yx=y.



thank you so much. can you also help me for… this one ?
– Kwangi Yu
2 days ago



@KwangiYu: You’re very welcome. I’ll take a look.
– Brian M. Scott
2 days ago

This sequence is known under the name “arithmetic-harmonic mean”. See (Proof the the Arithmetic-Harmonic Mean is expressible as the Geometric Mean) where it is proven that these sequences have a common limit which is the geometrical mean of the initial terms: x=y=√x1y1x=y=\sqrt{x_1y_1}.