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.

Note that $(y^2-5)^2=5-y$. It is also clear that $ y^2 \geq 5$ and $0<y$.

We have $0=y^4-10y^2+y+20=(y^2-y-4)(y^2+y-5)$.

We have $y^2+y-5>0$, so $y^2-y=4$.

P.S. Incidentally, if you want to find $y$, it is the positive root of $y^2-y-4=0$, which is $\frac{1+\sqrt{17}}{2}$.

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, $$ \lim_{k\to\infty}a_k=\frac{1+\sqrt{17}}{2} $$