# How to show $\{a_n\}$ and $\{b_n\}$ converges to the same limit $\sqrt 2$?

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 $a=b=\sqrt2$

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$a_nb_n=$?   –  Did Feb 6 '13 at 7:31

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 $\sqrt2$.

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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.
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.