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

Question

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

Answer 1

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\:.\ $

Answer 2

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.