$a_n$ given, then evaluate $\lim_{n\to \infty}a_n$. If  $$a_n=\sqrt{4+\sqrt{4+3\sqrt{4+5\sqrt{4+7\sqrt{\cdots\sqrt{4+(2n-1)\sqrt{4+(2n+1)}}}}}}}$$for any natural number $n$,
then evaluate $\lim_{n\to \infty}a_n$.
Note that $a_1=\sqrt{4+\sqrt{4+3}}$ and $a_2=\sqrt{4+\sqrt{4+3\sqrt{4+5}}}$.
I don't have any good idea. I need your help.
 A: Define
$$
f(x)=\sqrt{4+x\sqrt{4+(x+2)\sqrt{4+(x+4)\sqrt{4+\dots}}}}
$$
then $f(x)^2=4+xf(x+2)$. This indicates we should look at $f(x)=x+2$.
Considering $f(x)=x+2$, we are lead to show inductively that
$$
x+2=\small\sqrt{4+x\sqrt{4+(x+2)\sqrt{4+\dots+\sqrt{4+(x+2k-4)\sqrt{4+(x+2k-2)(x+2k+2)}}}}}
$$

Define
$$
f_{k,x}(y)=\left\{\begin{array}{}
y&\text{if }k=0\\
\sqrt{4+xf_{k-1,x+2}(y)}&\text{if }k>0
\end{array}\right.
$$
unrolled, that is
$$
f_{k,x}(y)=\small \sqrt{4+x\sqrt{4+(x+2)\sqrt{4+\dots+\sqrt{4+(x+2k-4)\sqrt{4+(x+2k-2)y}}}}}
$$
Note that $f_{0,x}(x+2)=x+2$. Suppose that, for some $k\ge0$, $f_{k,x}(x+2k+2)=x+2$. Then
$$
\begin{align}
f_{k+1,x}(x+2k+4)
&=\sqrt{4+xf_{k,x+2}(x+2k+4)}\\
&=\sqrt{4+x(x+4)}\\
&=x+2
\end{align}
$$
Therefore, for all $k\ge0$, we have $f_{k,x}(x+2k+2)=x+2$.

We want to show that
$$
\lim_{k\to\infty}f_{k,x}(1)=x+2
$$
Note that
$$
\frac{f_{0,x+2k}(x+2k+2)}{f_{0,x+2k}(1)}=\frac{x+2k+2}{1}
$$
Suppose that, for some $j\ge0$,
$$
\frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}\le(x+2k+2)^{1/2^j}
$$
Then
$$
\begin{align}
\frac{f_{j+1,x+2k-2j-2}(x+2k+2)}{f_{j+1,x+2k-2j-2}(1)}
&=\sqrt{\frac{4+(x+2k-2j-2)f_{j,x+2k-2j}(x+2k+2)}
{4+(x+2k-2j-2)f_{j,x+2k-2j}(1)}}\\
&\le\sqrt{\frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}}\\
&\le(x+2k+2)^{1/2^{j+1}}
\end{align}
$$
Therefore, for all $j\ge0$,
$$
\frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}\le(x+2k+2)^{1/2^j}
$$
in particular, for $j=k$,
$$
\frac{f_{k,x}(x+2k+2)}{f_{k,x}(1)}\le(x+2k+2)^{1/2^k}
$$

Since $f_{k,x}(x+2k+2)=x+2$, we have
$$
(x+2)(x+2k+2)^{-1/2^k}\le f_{k,x}(1)\le (x+2)
$$
and by the squeeze theorem, we have
$$
\lim_{k\to\infty}f_{k,x}(1)=x+2
$$
as desired. Setting $x=1$, we get the answer to the question to be $3$.
