Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$(x_n)_{n\geqslant1}$ and $(y_n)_{n\geqslant1}$ are two real sequences such that $x_1 > 0$, $y_1 > 0 $, $x_{n+1} = \frac{1}{2}(x_n + y_n)$ and $\dfrac{2}{y_{n+1}} = \dfrac{1}{x_n} + \dfrac{1}{y_n}$.

Do these two sequences converge to the same limit? If yes, what is their limit?

share|cite|improve this question
up vote 3 down vote accepted

$$ \begin{align} x_{n+1}-y_{n+1} &=\frac12\left(x_n+y_n\right)-\frac{2x_ny_n}{x_n+y_n}\\ &=\frac12\frac{(x_n-y_n)^2}{x_n+y_n}\tag{1} \end{align} $$ If $x_n,y_n>0$, $\left|\frac{x_n-y_n}{x_n+y_n}\right|<1$, so $\frac12\frac{(x_n-y_n)^2}{x_n+y_n}<\frac12|x_n-y_n|$. Therefore, $$ 0\le x_{n+1}-y_{n+1}\le\frac12|x_n-y_n|\tag{2} $$ Thus, $$ |x_n-y_n|\le\frac1{2^{n-1}}|x_1-y_1|\tag{3} $$ Therefore, $\displaystyle\lim_{n\to\infty}x_n-y_n=0$. For $n>1$, $(1)$ guarantees that $x_{n}\ge y_n$. Furthermore, since $x_{n+1}$ and $y_{n+1}$ are means of $x_n$ and $y_n$, we have $$ y_n\le y_{n+1}\le x_{n+1}\le x_n\tag{4} $$ Therefore, $x_n$ is monotonically decreasing and $y_n$ is monotonically decreasing, and both are bounded. Thus, $$ \lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n\tag{5} $$ Furthermore, as I noted in a comment to Jonas Kgomo's answer, $$ x_{n+1}y_{n+1}=\frac12(x_n+y_n)\frac{2x_ny_n}{x_n+y_n}=x_ny_n\tag{6} $$ This easily leads to the conclusion that $$ \lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=\sqrt{x_1y_1}\tag{7} $$

share|cite|improve this answer

if we divide 1st one by x_n


then 2nd one multiply by y_n

$$\frac{y_n}{y_{n+1}} =\frac{1}{2}\left(1+\frac{y_n}{x_n}\right)$$

share|cite|improve this answer
it seems like they are different not converging to the same limit – Jonas12 Nov 13 '12 at 9:41
There is a mistake in second equation – Norbert Nov 13 '12 at 9:59
@Norbert: I think that I fixed it. – robjohn Nov 13 '12 at 10:06
It appears that $x_{n+1}y_{n+1}=x_ny_n$, so the product of the limits would be $x_1y_1$. – robjohn Nov 13 '12 at 10:07

This is a simplified version of robjohn's answer.

By the AM-Hm inequality we have $y_n \leq x_n$. Moreover

$$x_n = \frac{x_{n-1}+y_{n-1}}{2} \leq \frac{ x_{n-1}+x_{n-1}}{2}=x_{n-1}$$ and


This shows that $y_1 \leq y_2 \leq ..\leq y_n \leq ..\leq x_n \leq x_{n-1} \leq ... \leq x_1$$

Hence both sequences are monotonic and bounded thus convergent. Let

$$\lim x_n=l \,;\, \lim y_n =m \,.$$

Then taking the limit in $x_{n+1}=\frac{x_n+y_n}{2}$ we get $l=m$.

share|cite|improve this answer
In the first two inequalities, we need $n>2$ since it is possible that $x_1<y_1$. – robjohn Nov 13 '12 at 16:21
It is true, that $(2)$ and $(3)$ in my answer are mostly unnecessary and $(1)$ simply justifies the AM-HM inequality. However, I do like the result in $(7)$ saying that $$ \lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=\sqrt{x_1y_1} $$ – robjohn Nov 13 '12 at 16:52
@robjohn Yes, I really like that part.. I realized while I was typing that I kinda re-say what you were saying, tried to give you credit for that; and to be honest I didn't see how to get the limit :) – N. S. Nov 13 '12 at 19:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.