3

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.

Dhrubajyoti Bhattacharjee
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2 Answers2

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

Peter Foreman
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  • One thing that I would love to know: how did you guess(or got to know or what struck you) that $2$ would be the value of the original expression? Earlier I found a value for a similar expression: $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\sqrt{1+5\sqrt{\ldots}}}}}}$ . Seeing the pattern I proceeded as follows: $n=\sqrt{n^2}=\sqrt{1+(n^2-1)}=\sqrt{1+(n-1)(n+1)}=\sqrt{1+(n-1)\sqrt{(n+1)^2}}=\sqrt{1+(n-1)\sqrt{1+(n+1)^2-1}}=\sqrt{1+(n-1)\sqrt{1+n(n+2)}}=\sqrt{1+(n-1)\sqrt{1+n\sqrt{(n+2)^2}}}\ldots$ Then substituting $n=3,$ we get $3$ as the required result. It was somewhat obvious. So I could solve it! – Dhrubajyoti Bhattacharjee Jun 04 '20 at 08:27
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    I found the value numerically then worked backwards from there. But as I said starting from any $n$ produces a valid radical for $n$ which only corresponds to your expression in the case $n=2$. – Peter Foreman Jun 04 '20 at 08:29
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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 $\alpha=2$.

Pablo Herrera
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  • Yes but as shown in my answer for $a_0=2$ the final term is $O(n)$ so the error tends to zero. For no other $a_0$ should this be true. – Peter Foreman Jun 04 '20 at 09:10
  • @PeterForeman excuse me how do you calculate the error for a constant sequence in this case? do you mean $a_n$ grows $O(n)$? – Pablo Herrera Jun 04 '20 at 09:24
  • @PeterForeman I've checked the innermost square term of your sequence that is $O((n+2)^{\frac{1}{2^n}})$ but is approaching to $1$ not $0$, when $n \to \infty$. – Pablo Herrera Jun 04 '20 at 10:08
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    That's not exactly how the error compounds. Consider $\sqrt{a+h}$. For small $h$ this differs from $\sqrt{a}$ by $h/(2\sqrt{a})$ so this error is more realistically $O(n/2^n)\to0$. – Peter Foreman Jun 04 '20 at 10:43
  • Now I understand thank you. – Pablo Herrera Jun 04 '20 at 13:37