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 $\sum (u_{n+1}-u_n)$.

As I assumed, the statement is not clear. Only the first two lines are the hypotheses. I'll do it tomorrow:
$a_{n+1} = \tfrac{a_n+b_n}{2}, b_{n+1} = \sqrt{a_nb_n}\rightarrow$ (arithmetic mean > geometric mean)
$0<b<b_{1}<a_{1}<a,\quad 0<b<b_{1}<b_{2}<a_{2}<a_{1}<a, \quad \text{etc.}\rightarrow \left\{ \begin{array}{lcc}
             a_{n} \quad \text{is decreasing} \\
             \\ b_{n} \quad \text{is increasing}
             \end{array}
   \right.$
$\{a_{n}\}$ and $\{b_{n}\}$ monotones and bounded $\rightarrow$ They are convergent $\rightarrow \left\{ \begin{array}{lcc}
             \exists \lim_{n\to \infty}\{a_{n}\}=m \\
             \\ \exists \lim_{n\to \infty}\{a_{n}\}=n
             \end{array}
   \right.$
$a_{n+1}=\dfrac{a_{n}+b_{n}}{2}\rightarrow \lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\dfrac{a_{n}+b_{n}}{2}\rightarrow m=\dfrac{m+n}{2}$ 
$\rightarrow m=n\rightarrow \lim_{n\to \infty}\{a_{n}\}=\lim_{n\to \infty}\{b_{n}\}=m$
$a_{n}$ is increasing $\rightarrow a_{n}\geq m$
$c_{n}=\sqrt{a_{n}^{2}-b_{n}^{2}}\rightarrow c_{n}^{2}=(a_{n}-b_{n})(a_{n}+b_{n})$
$c_{n+1}=\sqrt{a_{n+1}^{2}-b_{n+1}^{2}}=\sqrt{\left( \frac{a_{n}+b_{n}}{2}\right)^{2}-(\sqrt{a_{n}b_{n}})^{2}}$ 
$\rightarrow c_{n+1}=\sqrt{\frac{a_{n}^{2}+2a_{n}b_{n}+b_{n}^{2}-4a_{n}b_{n}}{4}}=\dfrac{1}{2}\sqrt{(a_{n}-b_{n})^{2}}=\dfrac{1}{2}(a_{n}-b_{n})$ 
$\rightarrow c_{n+1}=\dfrac{1}{2}\dfrac{c_{n}^{2}}{(a_{n}+b_{n})}=\dfrac{1}{2}\dfrac{c_{n}^{2}}{2a_{n}}\leq \dfrac{c_{n}^{2}}{4m}\rightarrow c_{n+1}\leq \dfrac{c_{n}^{2}}{4m}$
You've understood?
EDIT:
It defines: $u_{n}=-2^{-n}\ln(c_{n})$
$\rightarrow u_{n+1}=-2^{-n-1}\ln(c_{n+1})$
$\rightarrow u_{n+1}=-2^{-n-1}\ln\left( \frac{c_{n}^{2}}{4a_{n+1}}\right)$
$\rightarrow u_{n+1}=-2^{-n-1}[2\ln(c_{n})-\ln(4a_{n+1})]=-2^{-n}\ln(c_{n})+2^{-n-1}\ln(4a_{n+1})$
$\rightarrow u_{n+1}=u_{n}+2^{-n-1}\ln(4a_{n+1})$
$\rightarrow u_{n+1}-u_{n}=2^{-n-1}\left[ \ln(4m)+\ln\left( \frac{4a_{n+1}}{4m}\right)\right]$
$\rightarrow u_{n+1}-u_{n}=2^{-n-1}\left[ \ln(4m)+\ln\left( \frac{a_{n+1}}{m}\right)\right]$;
we put: $a_{n+1}=\ln\left( \frac{a_{n+1}}{m}\right)\rightarrow \left\{ \begin{array}{lcc}
             a_{n} \quad \text{decreases} \\
             \\ a_{n}>0 \\
             \\ \lim_{n\to \infty}a_{n}=0
             \end{array}
   \right.$ 
$u_{n}-u_{0}=\sum^{n-1}_{k=0}(u_{k+1}-u_{k})$
$\rightarrow -2^{-n}\ln(c_{n})-[-2^{-0}\ln(c_{0})]=\sum^{n}_{k=0}\left[2^{-k-1}[\ln(4m)+\alpha_{k+1}]\right]$
$\rightarrow -2^{-n}\ln(c_{n})+\ln \sqrt{a-b}=\ln(4m)\sum^{n-1}_{k=0}\dfrac{1}{2^{k+1}}+\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
$\rightarrow 2^{-n}\ln(c_{n})=-\ln(4m)\cdot \dfrac{\frac{1}{2}\left( 1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}}+\ln \sqrt{a-b}-\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
$\rightarrow 2^{-n}\ln(c_{n})=-\dfrac{2^{n}-1}{2^{n}}\ln(4m)-\ln \dfrac{1}{\sqrt{a-b}}-2^{n}\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
$\rightarrow \ln(c_{n})=\ln(4m)-2^{n}\left( \ln \dfrac{1}{\sqrt{a-b}}+\ln 4m\right)-2^{n}\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
$\rightarrow \ln(c_{n})=\ln(4m)-2^{n}\ln \dfrac{4m}{\sqrt{a-b}}-2^{n}\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
Where $\sum^{\infty}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}=K$, since 
$\dfrac{\frac{\alpha_{k+1}}{2^{k+1}}}{\frac{\alpha_{k}}{2^{k}}}=\dfrac{\alpha_{k+1}}{2\alpha_{k}}<1$ (Quotient criterion)
$\rightarrow \beta_{n}=K-\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}, \quad \text{with}\quad \lim_{n\to \infty}\beta_{n}=0$
$\rightarrow \ln(c_{n})=\ln(4m)-2^{n}\ln \dfrac{4m}{\sqrt{a-b}}-2^{n}(K-\beta_{n})$
$\rightarrow \ln(c_{n})=\ln(4m)-2^{n}\left( \ln \frac{4m}{\sqrt{a-b}}+K\right)-2^{n}\beta_{n}$
Let $l=\ln \dfrac{4m}{\sqrt{a-b}}+K>0$:
$$c_{n}=4me^{-2^{n}l-2^{n}\beta_{n}}$$
