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. 
 A: 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 $\frac{1+\sqrt{17}}{2}$.
A: 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,
$$
\lim_{k\to\infty}a_k=\frac{1+\sqrt{17}}{2}
$$
A: It's not a series at all, just a recursively defined sequence. Notice the successive terms alternating in sign means by constructing the terms recursively you have for any infinitieth term a different expression for the next one.
Compare $\{1, -1, 1, -1, ...\}$ which does not converge.
So, the sequence can only have a limit if the values are the same regardless of the expression. To test that, though, you need a way to make sure the first time you test it the infinitieth value has one sign, and the second time it has the other sign.
You need to test two sequences for convergence:
$a_{k+1}=\sqrt{5+\sqrt{5-a_k}}$ and $a_{k+1}=\sqrt{5-\sqrt{5+a_k}}$.
Suppose they converge. Since you defined the outermost sign to always be positive, if the second of the two converges to $a$, take $\sqrt{5+a}$. If the limit of the 1st one is the same as $\sqrt{5+a}$, then it doesn't matter whether the sign at the limit is positive or negative, so the original sequence you defined converges.
