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 $\sum (u_{n+1}-u_n)$.
 A: Assuming you have knowledge of theta fuctions, use the parametrization
$$ a_n = m\theta_3(q^{2^{n+1}})^2,\;
 b_n = m\theta_4(q^{2^{n+1}})^2,\;
 c_n = m\theta_2(q^{2^{n+1}})^2$$
for some $q$ which comes from the identity $\theta_3(q)^4=\theta_4(q)^4+\theta_2(q)^4$. By power series expansion
$$ c_n = 4m(q^{2^n} +2q^{52^n}+q^{92^n}+\cdots)$$
and so the rest is comparatively easy.
Using the suggested approach, we get the following. Without loss of generality, let $m=1$, since the
AGM is homogeneous. Next, let $t_n := -\ln(c_n/4)/2^n$, (minus to make $t_n>0$). We know$$c_{n+1} = c_n^2/(4a_{n+1}) \quad \textrm{and}\quad
 \ln(a_{n+1})/2^{n+1} = t_{n+1} - t_n.$$
Using the inequalities $a_0 > a_1 > \cdots>1,\;$ we know that $t_0 <t_1< \dots\;$ is increasing sequence but $\ln(a_0)/2^{n+1}>t_{n+1}-t_n$ and so $t_n\to t$ for some $t>0$. Summing the telescoping series we get
$$t -t_n = \sum_{k>n} \ln(a_{k})/2^k  \quad \textrm{and}\quad 
\ln(a_0)/2^n >t -t_n > 0.$$
A: 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 $l=\ln \dfrac{4m}{\sqrt{a-b}}+K>0$:
$$c_{n}=4me^{-2^{n}l-2^{n}\beta_{n}}$$
A: 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 $\ell \ne 0$.
