<s>[INST] Limit of $y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}}}$ I would appreciate any help with this problem:
If 
$$y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}}}$$
Then how do I find $y^2 - y$?
I'm not sure whether this is an arithmetic or geometric series. 
 [/INST] Define
$$
a_0=0\quad\text{and}\quad a_{k+1}=\sqrt{5+\sqrt{5-a_k}}
$$
Show that for $k\ge1$, $\sqrt5\le a_k\le\sqrt{5+\sqrt5}$.
Initially, $0\le a_0=0\le5$.
Suppose that $0\le a_k\le 5$. Then $\sqrt5\le a_{k+1}\le\sqrt{5+\sqrt5}$.
$$
\begin{align}
\left|\,a_{k+1}-a_k\,\right|
&=\frac{\left|\,\left(5+\sqrt{5-a_k}\right)-\left(5+\sqrt{5-a_{k-1}}\right)\,\right|}{a_{k+1}+a_k}\\
&=\frac{\left|\,\sqrt{5-a_k}-\sqrt{5-a_{k-1}}\,\right|}{a_{k+1}+a_k}\\
&=\frac{\left|\,a_k-a_{k-1}\,\right|}{(a_{k+1}+a_k)\left(\sqrt{5-a_k}+\sqrt{5-a_{k-1}}\right)}\\
&\le\frac{\left|\,a_k-a_{k-1}\,\right|}{\left(\sqrt5+\sqrt5\right)\left(\sqrt{5-\sqrt{5+\sqrt5}}+\sqrt{5-\sqrt{5+\sqrt5}}\right)}\\
&\le\frac{\left|\,a_k-a_{k-1}\,\right|}{13}
\end{align}
$$
Thus, $a_k$ converges since
$$
\lim_{n\to\infty}\sum_{k=1}^n(a_k-a_{k-1})=\lim_{n\to\infty}a_n-a_0
$$
converges absolutely; that is,
$$
\sum_{k=1}^\infty\left|\,a_k-a_{k-1}\right|\le\frac{13}{12}\sqrt{5+\sqrt5}
$$
Set $a=\lim\limits_{k\to\infty}a_k$. Since $\sqrt{5+\sqrt{5-x}}$ is continuous for $x\le5$, we have that $a=\sqrt{5+\sqrt{5-a}}$, which means
$$
\begin{align}
0
&=a^4-10a^2+a+20\\
&=(a^2-a-4)(a^2+a-5)
\end{align}
$$
The roots of $a^2-a-4$ are $\frac{1\pm\sqrt{17}}{2}$ and the roots of $a^2+a-5$ are $\frac{-1\pm\sqrt{21}}{2}$. The only one that is between $\sqrt5$ and $\sqrt{5+\sqrt5}$ is $\frac{1+\sqrt{17}}{2}$. Therefore,
$$
\lim_{k\to\infty}a_k=\frac{1+\sqrt{17}}{2}
$$
</s>