# Evaluating the nested radical $\sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}}$.

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

## 4 Answers

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

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.

• What a beauty! :D – AD. Oct 19 '10 at 18:29
• I just followed a "possible duplicate" link here and saw that my answer is a "possible duplicate" of yours :-) (+1) – robjohn Jul 2 '12 at 13:26

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 $$\lim_{n\to\infty}a_{m,n}=m+1.$$

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

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 $$\alpha,\beta,\gamma,\delta$$

• Don't you already have deleted answer just like this one? – Ennar Aug 19 at 13:01
• @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. – Cesareo Aug 19 at 13:13
• 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? – Ennar Aug 19 at 15:07
• @Ennar The facts may be to our liking or not but the facts are the facts and we can't ignore it. – Cesareo Aug 19 at 15:16
• Nor prove it, apparently... – Ennar Aug 19 at 15:17

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