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How does one prove the following limit? $$ \lim_{n \to \infty} \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}} = 3. $$

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This is the special case $\rm\ x,\:n,\:a = 2,\:1,\:0\ $ in Ramanujan's second notebook, chapter XII, entry 4:

$$\rm x + n + a\ =\ \sqrt{ax + (n+a)^2 + x\sqrt{a(x+n) + (n+a)^2 + (x+n) \sqrt{\cdots}}} $$

Below is Ramanujan's solution of the given special case - which was submitted to a journal in April 1911. Note that his solution is incomplete (exercise: why?). For further discussion see this 1935 Monthly article, Herschfeld: On infinite radicals. It also appeared as Problem A6 on the 27th Putnam competition, 1966. Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $\ \sqrt{a_1 + \sqrt{a_2 +\:\cdots\: +\sqrt{a_n}}}\ \ $ is that $\rm\displaystyle\ \ {\overline \lim}_{n\to\infty}\frac{\log{a_n}}{2^n}\ < \infty\:.\ $

alt text

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    $\begingroup$ The above solution is from which journal? Can you give me the link? $\endgroup$ – Pragyaditya Das May 6 '17 at 16:01
  • $\begingroup$ @PragyadityaDas: the image is from Collected Papers of Ramanujan. $\endgroup$ – Paramanand Singh Aug 18 '18 at 18:46
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This is Ramanujan's famous nested radical.

More information can be found here: http://www.isibang.ac.in/~sury/ramanujanday.pdf

See Also: http://mathworld.wolfram.com/NestedRadical.html (number 26).

Apparently, this is how he came up with it (sorry, no reference for this claim).

Start with

$$3 = \sqrt{9} = \sqrt{1 + 8} = \sqrt{1 + 2 \cdot 4}$$ $$ = \sqrt{1 + 2\sqrt{16}} = \sqrt{1 + 2\sqrt{1 + 3 \cdot 5}}$$ $$ = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25}}} = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{1 + 4 \cdot 6}}}$$ etc.

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    $\begingroup$ What a beauty! :D $\endgroup$ – AD. Oct 19 '10 at 18:29
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    $\begingroup$ I just followed a "possible duplicate" link here and saw that my answer is a "possible duplicate" of yours :-) (+1) $\endgroup$ – robjohn Jul 2 '12 at 13:26
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Let me provide a full and simple proof here (6 years later)

Set, for $m<n$ $$ a_{m,n}=\sqrt{1+m\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+n}}}}. $$ It is can be shown that $$a_{m,n}<m+1,\tag{1}$$ in the following way. (Backwards induction.) First, $a_{n,n}=\sqrt{1+n}<1+n$, and if $a_{k+1,n}<k+2$, then $$ a_{k,n}=\sqrt{1+ka_{k+1,n}}<\sqrt{1+k(k+2})=\sqrt{k^2+2k+1}=k+1. $$ In particular, in the same way we can show that $$ m+1=\sqrt{1+m\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+{\color{red}{n^2+2n}}}}}} $$ Next, observe that $$ 0<m+1-a_{m,n}= \\ =\sqrt{1+m\sqrt{1+\cdots+(n-1)\sqrt{1+{\color{red}{n^2+2n}}}}} -\sqrt{1+m\sqrt{1+\cdots+(n-1)\sqrt{1+{n}}}} \\ =\frac{m\Big(\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+{\color{red}{n^2+2n}}}}} -\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+{n}}}}\Big)}{\sqrt{1+m\sqrt{1+\cdots+(n-1)\sqrt{1+{\color{red}{n^2+2n}}}}} +\sqrt{1+m\sqrt{1+\cdots+(n-1)\sqrt{1+{n}}}}} \\ <\frac{m}{m+2}(m+2-a_{m+1,n}) <\cdots < \frac{m(m+1)\cdots (n-1)}{(m+2)(m+3)\cdots(n+1)}(n+1-a_{n,n})=\frac{m(m+1)(n+1-\sqrt{n+1})}{n(n+1)}<\frac{m(m+1)}{n}. $$ Thus $$ \lim_{n\to\infty}a_{m,n}=m+1. $$

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    $\begingroup$ Lol, 6 years later indeed. $\endgroup$ – Simply Beautiful Art Dec 6 '16 at 22:14
  • $\begingroup$ Exactly what I did when I encountered this problem! Nicely done. $\endgroup$ – Stelios Sachpazis Aug 16 at 18:19
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Considering that

$$ f(x) = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{\cdots}}}} $$

or

$$ f(x) = \sqrt{1+x f(x+1)} $$

or

$$ f^2(x) = 1+xf(x+1) $$

this gives us the possibility of calculating solutions with the structure $f(x) = a x + b$ or equating

$$ (ax+b)^2 = 1+x(a(x+1)+b) $$

or

$$ (1-b^2)+(a+b-2ab)x+a(a-1)x^2 = 0\;\;\forall x \Rightarrow a = b= 1 $$

hence

$$ f(2) = 3 $$

is a solution.

NOTE

Thus can be solved any recurrence relationship of type

$$ f(x) = \sqrt{\alpha +\beta x f(\gamma x + \delta)} $$

for suitable values of $\alpha,\beta,\gamma,\delta$

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  • $\begingroup$ Don't you already have deleted answer just like this one? $\endgroup$ – Ennar Aug 19 at 13:01
  • $\begingroup$ @Ennar The deleted answer is not like this one. I usually learn from my mistakes. Now it seems to me that the result is justified. $\endgroup$ – Cesareo Aug 19 at 13:13
  • $\begingroup$ Let's ignore that it is not clear why $f$ is well defined at all (it can be shown by considering appropriate increasing, bounded above sequence). However, there is absolutely no reason that $f$ should be linear. Take any function defined on $(0,1]$ and then use your recurrence to extend it on $(0,\infty)$. How do you know your $f$ is not something more exotic? $\endgroup$ – Ennar Aug 19 at 15:07
  • $\begingroup$ @Ennar The facts may be to our liking or not but the facts are the facts and we can't ignore it. $\endgroup$ – Cesareo Aug 19 at 15:16
  • $\begingroup$ Nor prove it, apparently... $\endgroup$ – Ennar Aug 19 at 15:17