How does one prove the following limit? $$ \lim_{n \to \infty} \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}} = 3. $$
Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $.
3 Answers
Let me provide a full and simple proof here (6 years later)
Set, for $m
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4Lol, 6 years later indeed. – 2016-12-06
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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13What a beauty! :D – 2010-10-19
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2I just followed a "possible duplicate" link here and saw that [my answer](http://math.stackexchange.com/a/165693) is a "possible duplicate" of yours :-) (+1) – 2012-07-02
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\:.\ $
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1The above solution is from which journal? Can you give me the link? – 2017-05-06
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0@PragyadityaDas: the image is from Collected Papers of Ramanujan. – 2018-08-18