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

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

• In this video youtu.be/r5BGIi84arY by MindYourDecisions explains the "Ramanujan's radical brain teaser".
– user845875
Commented Jul 30, 2021 at 18:23
• Also check out his blog post mindyourdecisions.com/blog/2016/05/01/…
– user845875
Commented Jul 30, 2021 at 18:26
• $$\frac{1}{n+1}\left(\frac{1}{n-1}-\frac{1}{n-1}\left(...\left(\frac{1}{3}-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{2}x^{2}\right)^{2}\right)^{2}\right)...\right)^{2}-1=f(x)$$ $f(3)=n$ by induction then use the fact $f(f^{-1}(n))=n$ now take the limit . Commented Jul 3 at 14:25

## 3 Answers

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 Commented Oct 19, 2010 at 18:29
• I just followed a "possible duplicate" link here and saw that my answer is a "possible duplicate" of yours :-) (+1)
– robjohn
Commented Jul 2, 2012 at 13:26
• According to this reasoning,$$4=\sqrt{16}=\sqrt{1+2\sqrt{\frac{225}{4}}}=\sqrt{1+2\sqrt{1+3\sqrt{...}}}$$. This should go on forever. If I accept your reasoning it implies that I also accept this reasoning. I think you had better prove it by some other technique.
– user730361
Commented Nov 26, 2020 at 14:23

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? Commented May 6, 2017 at 16:01
• @PragyadityaDas: the image is from Collected Papers of Ramanujan. Commented Aug 18, 2018 at 18:46

Let me provide a full and simple proof here (6 years later)

Set, for $$m $$a_{m,n}=\sqrt{1+m\sqrt{1+(m+1)\sqrt{1+\cdots+(n-1)\sqrt{1+n}}}}.$$ We shall first show that $$a_{m,n} in the following way using Backwards induction. Clearly, $$a_{n,n}=\sqrt{1+n}<1+n$$, and if $$a_{k+1,n}, for some $$k, 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 $$0 Thus $$\lim_{n\to\infty}a_{m,n}=m+1.$$

• Lol, 6 years later indeed. Commented Dec 6, 2016 at 22:14
• Exactly what I did when I encountered this problem! Nicely done. Commented Aug 16, 2019 at 18:19
• this is beautiful. The original proof never sat well with me, it more or less amounted to knowing the solution beforehand. This is a nice forward proof. +1 Commented May 5, 2020 at 2:05