Find the value of $\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\cdots}}}}}$ .

Question

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.

Answer 1

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 $$3=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\cdots}}}}$$ $$4=\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}$$

Answer 2

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 $\alpha=2$.