Let $a$ and $b$ reals with $a > b > 0$. Let $(a_n)$ and $(b_n)$ with $a_0 = a$, $b_0 = b$ and

$$a_{n+1} = \tfrac{a_n+b_n}{2} \quad\text{;}\quad b_{n+1} = \sqrt{a_nb_n}$$

We know that $\lim_{n \rightarrow +\infty} a_n = \lim_{n \rightarrow +\infty} b_n = m$.

With $c_n = \sqrt{a_n^2-b_n^2}$, we know that $\lim_{n \rightarrow +\infty} c_n =0$ and that $c_{n+1} \leq \frac{c_n^2}{4m}$. It has already been proved that.

$$ a_n-m = \sum_{k=n+1}^{+\infty} c_k$$

The question is to prove that $a_n-m \sim \frac{c_n^2}{4m}$ when $n \rightarrow +\infty$.


Hint: $\dfrac{c_n^2}{4m(a_n-m)}=\dfrac{a_n^2-b_n^2}{4m(a_n-m)}=(\dfrac{a_n}{4m}+\dfrac{1}{4})+\dfrac{m-b_n}{a_n-m}(\dfrac{b_n}{4m}+\dfrac{1}{4}).$

So what remains to be shown now is $$\lim_{n\to\infty}\dfrac{m-b_n}{a_n-m}=1,$$ which should be seen easily.

  • $\begingroup$ Thanks for help, but is the final limit really simple to prove ? $\endgroup$ – MB4E Aug 29 '17 at 13:48
  • $\begingroup$ sorry i think its harder then i thought. I will let u know if got something. $\endgroup$ – dezdichado Aug 29 '17 at 23:55

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