Find the value of $\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\cdots}}}}}$ . We have:
$\sqrt{(1+0)+\sqrt{(4+1)+\sqrt{(9+2)+\sqrt{(16+3)+\sqrt{(25+4)+\cdots}}}}}.$
Basically I'm not getting any clue at the moment for reducing the infinite nested radicals. Any hint would be helpful. Thanks in advance.
 A: Note that we have the identity
$$n=\sqrt{(n^2-n-1)+(n+1)}$$
Which we can apply indefinitely to give
\begin{align}
2
&=\sqrt{1+3}\\
&=\sqrt{1+\sqrt{5+4}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+5}}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+6}}}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+7}}}}}\\
\end{align}
Note that the $n$th line above differs from the provided expression by an $O(n)$ term in the innermost square root. Due to $n$ square roots this error is reduced to zero as $n\to\infty$.
Edit: As shown above, ignoring some of the first terms gives radical expressions for every natural number. For example
$$3=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\cdots}}}}$$
$$4=\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}$$
A: The answers should be ambiguous. Here infinite is a problem. There are infinite numbers where you can make this construction. Observe that the nested radical satisfies $a_n=\sqrt{n²-n+1+a_{n+1}}$. So if we start with $a_0=3$. You can calculate $a_1,a_2,...,$ and so on.
\begin{align}
3 &= \sqrt{1+8}\\
&=\sqrt{1+\sqrt{5+59}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+3474}}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+12068657}}}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+145652481783620}}}}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+...}}}}}
\end{align}
It would be a interest question for which $a_0<\alpha$ this algorithm fails in finite steps. For example if $a_0=3/2$ this fails for $n=5$ with $a_5=-\frac{1201503}{65536}$. We need to guarantee certain growing. I conjecture the critical value is $\alpha=2$. 
