Note: I would like to add this is very similar to this problem posted a couple months ago, which is the inspiration for this question, was proven to converge in one of the answers, and I believe a similar technique could be used to show that this infinitely nested radical converges as well.
My first intuition to solve this was to put this in terms of $n$ using $$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$ So then we have $$\sqrt{\frac{1(1+1)}{2}+\sqrt{\frac{2(2+1)}{2}+\sqrt{\frac{3(3+1)}{2}..}}}$$ From here it should be simple to define a recursive relationship like this one $$a_{n} = \sqrt{\frac{n(n+1)}{2}+a_{n+1}}$$ Where $a_{1} = \sqrt{\frac{1(1+1)}{2}+a_{2}}$ and then it is clear that $$a_{n}^{2} - a_{n+1} = \frac{n(n+1)}{2}$$ I have no idea where to go from here.
Edit: as @mattapow pointed out in the comments, we can attempt to solve this using Ramanujans's Infinite Radical Technique but I believe this runs into some problems. As stated above there is a clear recursive relation: $$a_{n}^2 = \frac{n(n+1)}{2} + a_{n+1}$$ Ramanujans's Infinite Radical Technique involves writing this in the form $$F(x)^2 = p(x) + F(x+1)$$ and finding an 'ansatz', or essentially solving for $F(x)=ax+b$ in our case by equating coefficients. Here is my attempt, and inevitable failure: $$F(x)^2 - F(x+1) = \frac{x(x+1)}{2}$$ $$(ax+b)^2 - a(x+1)-b = \frac{1}{2}x^2 + \frac{1}{2}x$$ $$a^2x^2+2abx+b^2-ax-a-b=\frac{1}{2}x^2 + \frac{1}{2}x$$ $$a^2x^2+(2ab-a)x -a-b+b^2=\frac{1}{2}x^2 + \frac{1}{2}x$$ We can now equate coefficients to get a system of equations: $$a^2=\frac{1}{2}; 2ab - a = \frac{1}{2}; -a-b+b^2 = 0$$ Which has no solutions. I cannot think of another approach.
from Hot Weekly Questions - Mathematics Stack Exchange
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