Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $a_1=2$ and $b_1=1$ and for $n \geq 1$ , $a_{n+1}=\dfrac{a_n+b_n}{2}, b_{n+1}=\dfrac{2a_nb_n}{a_n+b_n}$.Show that the sequences $\{a_n\}$ and $\{b_n\}$ converges to the same limit $\sqrt 2$.

trial: Let $a_n$ converges to $a$ and $b_n$ converges to $b$. So I get $a=\dfrac{a+b}{2}, b=\dfrac{2ab}{a+b}$ and $a=b$ But how I show $a=b=\sqrt2$

share|improve this question
$a_nb_n=$? $ $ $ $ –  Did Feb 6 '13 at 7:31

3 Answers 3

up vote 4 down vote accepted

First note that $$a_{n+1} \geq b_{n+1} \,\,\,\,\,\, (\text{Since Arithmetic mean is greater than the Harmonic mean})$$ Further, $$a_{n+1} - b_{n+1} = \dfrac{(a_n-b_n)^2}{2(a_n+b_n)}$$ Now since $2 \geq a_n>b_n \geq 1$, we have $a_n + b_n > 2$ and hence $$a_{n+1} - b_{n+1} = \dfrac{(a_n-b_n)^2}{2(a_n+b_n)} < \dfrac{(a_n-b_n)^2}4 < \dfrac{(a_1-b_1)^2}{4^{2n-1}} = \dfrac1{4^{2n-1}} \,\,\,\,\,\, (\spadesuit)$$ Hence, since $a_n$ is a bounded monotone decreasing sequence and $b_n$ is a bounded monotone increasing sequence both converge and from $ (\spadesuit)$, we get that $$a_n \to b_n$$Further, $$a_{n+1} b_{n+1} = a_nb_n$$ and hence $$a_n b_n = a_1 b_1 = 2$$ Hence, conclude that both should converge to $\sqrt2$.

share|improve this answer

HINT: Show first that $a_nb_n=2$ for all $n$. Then show that $a_n>a_{n+1}>b_{n+1}>b_n$ for all $n$. The AM-GM inequality may be helpful.

share|improve this answer

Hint: Show that the limit of $\frac{a_{n+1}}{b_{n+1}}$ is $1$ when $n\to \infty$ to see that they are asymptotically convergent.

share|improve this answer
Nice observation!+1 –  amWhy Feb 7 '13 at 3:14

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.