Asymptotics of a coupled sequence In this paper, the authors make the passing remark "simple analysis reveals that" the coupled sequence

$\mu_k = \theta_k \left(\theta_{k-1}^{-1} - 1\right)$

$\theta_{k+1} = \frac{\sqrt{\theta_k^4 + 4\theta_k^2} - \theta_k^2}{2}$

is such that $\mu_k$ asymptotically equals $1 - \frac{3}{k} + \mathcal{O}(1/k^2)$, but I have been unable to prove this. I was wondering if anyone could point me in the right direction as to how I would show something like this?
This question is very similar to MSE question 2861768 and several
other questions.
The idea from these questions is to use the Stolz theorem.

For simplicity, define
$\,f(x) := (\sqrt{4x^2+x^4}-x^2)/2.\,$ Define a sequence
$\,a_{n+1} = f(a_n)\,$ where $\,a_1&gt;0.\,$
Next, define
 $$ b_n := 1/a_n, \quad\text{and} \quad
  e(x) := \frac12 + x\sqrt{1+\frac1{4x^2}} \tag{1}$$ 
where
$\, b_{n+1} = e(b_n)\,$ and
$\,\lim_{ x\to \infty} e(x)-x = \frac12.\,$ By the Stolz theorem
we get $\, b_n = \frac{n}2 + O(1)\,$ and thus,
$\, a_n = \frac2{n} + O(n^{-2}).$

The inverse function of $\,f(x)\,$ is $\,f^{(-1)}(x) = x/\sqrt{1-x},\,$ and we get

$$ \mu_n := a_n\!\left(\!\frac1{a_{n-1}}
   \!-\! 1 \!\right)\!=\!
   a_n\!\left(\!\frac{\sqrt{1\!-\!a_n}}{a_n}
   \!-\! 1\!\right) \!=\! 1 \!-\! \frac32 a_n
   \!+\! O(a_n^2) \tag{2}$$
but since $\,a_n = 2/n + O(n^{-2})\,$ now
$\, \mu_n = 1 - 3/n + O(n^{-2}) \,$ which is what we wanted to prove.

ADVANCED results (only if you are interested):

Define $\,x := 1/n, \, y := \ln(x).\,$
For some constant $\,c\,$ only depending on $\,a_1,\,$
for example, if $\,a_1=1\,$ then $\,c \approx -1.2929418036,\,$
we get an expansion of $\,a_n = g(1/n) = h(2n-c-\ln(2))\,$ where $$  g(x) \approx
 2\,x + (c+y)\,x^2 +((c+c^2) + (1+2c)y + y^2)\,x^3/2 + O(x^4) \tag{3}$$
and
$$ h(x) \approx 4 / (x + \ln(x + \ln(x + \frac16 + \ln(x + \frac{29}{72} + \ln(x + \frac{437}{640} + \dots))))). \tag{4}$$ 

The expansion $(3)$ implies that
$$ a_{n+1}(1/a_n - 1) = 1 - 3\,x + (-1-3c-3y)\,x^2/2 + O(x^3). \tag{5}$$

Notice that $\,\frac1{n+1}=\frac{x}{x+1}\,$ and thus the recursion for $\,a_n\,$ is equivalent to $\,g(\frac{x}{1+x}) = f(g(x))\,$ and from which the coefficients of the series expansion of $\,g(x)\,$ can be solved for term by term.

MSE question 3215 has an answer to the question of the convergence rate of $\,a_n\,$ for general $\,f(x).\,$
This question is very similar to MSE question 2861768 and several
other questions.
The idea from these questions is to use the Stolz theorem.

For simplicity, define
$\,f(x) := (\sqrt{4x^2+x^4}-x^2)/2.\,$ Define a sequence
$\,a_{n+1} = f(a_n)\,$ where $\,a_1&gt;0.\,$
Next, define
 $$ b_n := 1/a_n, \quad\text{and} \quad
  e(x) := \frac12 + x\sqrt{1+\frac1{4x^2}} \tag{1}$$ 
where
$\, b_{n+1} = e(b_n)\,$ and
$\,\lim_{ x\to \infty} e(x)-x = \frac12.\,$ By the Stolz theorem
we get $\, b_n = \frac{n}2 + O(1)\,$ and thus,
$\, a_n = \frac2{n} + O(n^{-2}).$

The inverse function of $\,f(x)\,$ is $\,f^{(-1)}(x) = x/\sqrt{1-x},\,$ and we get

$$ \mu_n := a_n\!\left(\!\frac1{a_{n-1}}
   \!-\! 1 \!\right)\!=\!
   a_n\!\left(\!\frac{\sqrt{1\!-\!a_n}}{a_n}
   \!-\! 1\!\right) \!=\! 1 \!-\! \frac32 a_n
   \!+\! O(a_n^2) \tag{2}$$
but since $\,a_n = 2/n + O(n^{-2})\,$ now
$\, \mu_n = 1 - 3/n + O(n^{-2}) \,$ which is what we wanted to prove.

ADVANCED results (only if you are interested):

Define $\,x := 1/n, \, y := \ln(x).\,$
For some constant $\,c\,$ only depending on $\,a_1,\,$
for example, if $\,a_1=1\,$ then $\,c \approx -1.2929418036,\,$
we get an expansion of $\,a_n = g(1/n) = h(2n-c-\ln(2))\,$ where $$  g(x) \approx
 2\,x + (c+y)\,x^2 +((c+c^2) + (1+2c)y + y^2)\,x^3/2 + O(x^4) \tag{3}$$
and
$$ h(x) \approx 4 / (x + \ln(x + \ln(x + \frac16 + \ln(x + \frac{29}{72} + \ln(x + \frac{437}{640} + \dots))))). \tag{4}$$ 

The expansion $(3)$ implies that
$$ a_{n+1}(1/a_n - 1) = 1 - 3\,x + (-1-3c-3y)\,x^2/2 + O(x^3). \tag{5}$$

Notice that $\,\frac1{n+1}=\frac{x}{x+1}\,$ and thus the recursion for $\,a_n\,$ is equivalent to $\,g(\frac{x}{1+x}) = f(g(x))\,$ and from which the coefficients of the series expansion of $\,g(x)\,$ can be solved for term by term.

MSE question 3215 has an answer to the question of the convergence rate of $\,a_n\,$ for general $\,f(x).\,$