<s>[INST] Arithmetic-geometric mean, prove that $c_n = 4me^{-\ell 2^n+\epsilon_n}$ 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 $\lim_{n \rightarrow +\infty} a_n = \lim_{n \rightarrow +\infty} b_n = m$.
Let $c_n = \sqrt{a_n^2-b_n^2}$.
We know that $\lim_{n \rightarrow +\infty} c_n =0$ and that $c_{n+1} \leq \frac{c_n^2}{4m}$.
The question is to prove that there is $\ell > 0$ and $\epsilon_n > 0$ with $\lim_{n \rightarrow +\infty} \epsilon_n = 0$ such as
$$c_n = 4me^{-\ell 2^n+\epsilon_n}$$
An indication is to use $u_n = -2^{-n} \ln(c_n)$ and $\sum (u_{n+1}-u_n)$.
 [/INST] Using $c_{n+1} = \frac{c_n^2}{2(a_n+b_n)} = \frac{c_n^2}{4a_{n+1}}$, we get.
\begin{align*}
u_{n+1} & = -2^{-n-1} \ln(c_{n+1}) \\
& = -2^{-n-1} (\ln(c_n)-\ln(4a_{n+1})) \\
& = u_n+2^{-n-1} \ln(4a_{n+1})
\end{align*}
$\ln(4a_{n+1}) = \ln(4m)+\underbrace{\ln(\frac{4a_{n+1}}{4m})}_{\alpha_n}$. The sequence $(\alpha_n)$ is positive and decreasing to zero.
$$ u_n-u_0 = \sum_{k=0}^{n-1} \left( u_{k+1}-u_k \right) = \sum_{k=0}^{n-1} 2^{-k-1} \left( \ln(4m)+\alpha_k \right) $$
Since $\sum_{k=0}^{n-1} 2^{-k-1} = 1-2^{-n}$, we get.
$$ u_n = u_0+\ln(4m)-2^{-n} \ln(4m)+\sum_{k=0}^{n-1} 2^{-k-1} \alpha_k $$
Since $\sum_{k=0}^{n-1} 2^{-k-1} \alpha_k \leqslant \sum_{k=0}^{n-1} 2^{-k-1} \alpha_0 \leqslant \alpha_0 \left( 1-2^{-n} \right)$, we get.
$$ \sum_{k=0}^{+ \infty} 2^{-k-1} \alpha_k = S \quad\text{with}\quad 0 < S \leqslant \alpha_0 $$
Now, the rest.
$$ R_n = \sum_{k=n}^{+\infty} 2^{-k-1} \alpha_k \leqslant \alpha_n \sum_{k=n}^{+\infty} 2^{-k-1} \leqslant 2^{-n}\alpha_n $$
Finally.
\begin{align*}
u_n & = u_0+\ln(4m)-2^{-n} \ln(4m)+S-R_n \\
& = u_0+\ln(4m)+S-2^{-n} \left( \ln(4m)+2^nR_n \right)
\end{align*}
Let $\ell = u_0+\ln(4m)+S$ and $\varepsilon_n = 2^nR_n$.
The last problem is to prove that $\ell > 0$.
As $u_n = -2^{-n} \ln(c_n)$ and $\lim_{n \rightarrow +\infty} c_n = 0$, $u_n > 0$ after some rank and so $\lim_{n \rightarrow +\infty} u_n = \ell \geq 0$. If $\ell = 0$, $c_n = 4me^{\varepsilon_n}$ and $\lim_{n \rightarrow +\infty} c_n = 4m \ne 0$. So $\ell \ne 0$.
</s>