Prove that a periodic nested radical converges to the largest real root of the corresponding polynomial Let $a_k$ be a sequence of fixed positive integers, $k \in [1,n]$. Consider the periodic nested radical:
$$x=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n+x}}}$$
We can transform this nested radical into the polynomial equation:
$$\left( \cdots \left(\left(x^2-a_1 \right)^2-a_2\right)^2-\cdots-a_{n-1}\right)^2-x-a_n=0$$
This equation has order $2^n$, so it has exactly $2^n$ roots.
I observed that the nested radical converges to the largest positive real root of this equation (for small $n$ at least).

How to prove that this kind of equation will always have a positive real root, and that the nested radical converges to the largest of them? (If this is true)

It's obvious that the nested radical should converge to a real and positive value.

Edit
An example. Consider the periodic nested radical, satisfying:
$$x=\sqrt{3+\sqrt{5+\sqrt{2+\sqrt{7+x}}}}$$
It's value can be found by iterations of the above expression:
$$x \approx 2.38592$$
Now, consider the polynomial equation, obtained from this radical:
$$\left(\left(\left(x^2-3 \right)^2-5\right)^2-2\right)^2-x-7=0$$
It has $16$ roots, exactly $8$ real roots and $8$ complex roots.
Of the real roots there are $4$ positive roots and $4$ negative roots.
Four positive roots are:
$$x_1 \approx 0.566147$$
$$x_2 \approx 1.15226$$
$$x_3 \approx 2.15887$$
$$x_4 \approx 2.38592$$
 A: I know this is not a formal answer but consider a polynomial $x^2 -c = 0$. This has 2 solutions $\sqrt{c}$ and $-\sqrt{c}$.
When you convert this polynomial equation into a fixed point equation by taking square roots, you will obtain the following equation
$x = \sqrt{c}$.
This second equation has only one solution, which is the positive one because of the square root, $\sqrt{c}$.
Again, now consider the equation $x = \sqrt{c + x}$. (Eq 1)
Take squares of both sides, we have $x^2 = c + x \implies x^2 - x - c = 0$.
$\Delta = 1 + 4c$
$x_1 = \frac{1 + \sqrt{1 + 4c}}{2}$ and $x_2 = \frac{1 - \sqrt{1 + 4c}}{2}$.
We know $\sqrt{1+4c} > 1$ thats why the $x_2$ root of the polynomial is not the solution of the original equation (Eq 1).
Thus, when you take squares of the original $x = \sqrt{a_1 + \sqrt{a_2 + ...\sqrt{a_n + x}}}$ equation, each time you will double the number of solutions with contributions of corresponding (Eq 1) omitted $-$ signed root, while the original solution was the bigger one, which is obtained by $+$ sign in discriminant.
A: Here is a partial solution ; I can
show that the limit always exists, but I cannot show (yet) that
the limit is independent of the initial value (this would in turn immediately
yield the fact that the limit is the largest root).
Let $f(x)=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n+x}}}$
and $P(x)=((x^2-a_1)^2-a_2)^2-\cdots-a_{n-1})^2-a_n$. Then $P$ 
is a polynomial of degree $2^n$, $f$ and $P$ are (formally at least) inverses of each other, so that $f(x)=y$ forces $P(y)=x$, though the converse is not true. 
In fact, unnesting inequalities as in the OP and taking care of signs , we have
$$
f(x)=y \Leftrightarrow \Big( P(y)=x, \text{and} \ y \geq f(0) \Big) \tag{1}
$$
In particular,
$$
f(x)=x \Leftrightarrow \Big( P(x)=x, \text{and} \ x \geq f(0) \Big) \tag{2}
$$
Since the square root function and all the $x\mapsto x+a$ function 
are increasing, $f$ which is a big composite of all such functions is
increasing also.
Consider the sequence $u=(u_p)$ defined by $u_0=0$, $u_{p+1}=f(u_p)$.
Then $u_0<u_1$, so $f(u_0)<f(u_1)$, i.e. $u_1<u_2$ and by induction we see
that $u$ is increasing.
It follows that $u$ either converges to a finite limit or diverges to $+\infty$.
But if it diverged to $+\infty$, $u_p=P(u_{p+1})$ would be asymptotically
equivalent to $u_{p+1}^{2^n}$ contradicting $u_p<u_{p+1}$ for large enough
$p$.
So $u$ must converge to a finite limit $l$. By (2) above, $l$ must be
a root of $P$. Let $m$ be any value $>l$. Then $m>u_k$ for every $k$,
hence $f(m)>f(u_k)=u_{k+1}$. Passing to the limit, we have $f(m)\geq l$.
Consider the sequence $v=(v_p)$ defined by $v_0=m$, $v_{p+1}=f(v_p)$.
By induction, we have $v_p\geq l$ for every $p$. Also, $v_p=P(v_{p+1})$ by (1). By arguments already used above, $v$ is monotonic and cannot diverge to $+\infty$. So $v$ must have a finite limit.
(... Proof to be finished ; looks promising from here ...)
