# Solving the infinite radical $\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+...}}}}$

$$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}}$$

This is a modification on the well-known Ramanujan infinite radical, $$\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots}}}}$$, except it cannot be solved by the conventional method -- the functional equation $$F(x)^2=ax+(n+a)^2+xF(x+n)$$, since setting $$n=1$$ with $$a=0$$ requires having $$(n+a)^2=1$$, not $$6$$.

Here are some alternative methods I've tried:

• The functional equation we have instead for this infinite radical is $$F(x)^2=6+xF(x+1)$$. I've tried to solve this, but unfortunately it's easy to demonstrate that $$F(x)$$ cannot be a simple linear function $$F(x)=ax+b$$. I've tried some slightly more complicated versions -- the equation for a hyperbola, etc. -- but nothing seems to work.
• I've tried factoring stuff out from the radical to bring it to a more tenable form. Perhaps not a satisfactorily rigorous approach, I thought of factoring out $$\sqrt{6^{N/2}}$$ where $$N\to\infty$$, which allows us to transform the radical into $$6^{-N/2}\sqrt{6^{N+1}+\sqrt{6^{2N+1}+2\sqrt{6^{4N+1}+\cdots}}}$$, which can be treated as having each term a power of $$6^{N/2}$$ in the limit. For a radical of the form $$\sqrt{\alpha^2+\sqrt{\alpha^4+2\sqrt{\alpha^8+\cdots}}}$$ we have the functional equation $$F(x)^2=\alpha^{2^x}+xF(x+1)$$, or upon letting $$F(x)=\alpha^{2^x}p(x)$$, you get $$p(x)^2-xp(x+1)=\alpha^{-2^x}$$, but I'm stuck there.
• Similarly, I tried factoring out some arbitrary $$N$$ then factoring out a term from each radical inside such that the coefficients go from being $$1,2,3,\cdots$$ to a constant $$1/N,1/N,1/N...$$, transforming the radical into $$N\sqrt{\frac6{N^2}+\frac1N\sqrt{\frac6{N^2}+\frac1N\sqrt{\frac{24}{N^2}+\frac1N\sqrt{\frac{864}{N^2}+\frac1N\sqrt{\frac{1990656}{N^2}+\cdots}}}}}$$ where the added terms go as $$k_1=6$$, $$k_{n+1}=\frac{n^2}6k_n^2$$. But how might one proceed?
• I considered differentiating the function $$G(x)=\sqrt{x+\sqrt{x+2\sqrt{x+3\sqrt{x+\cdots}}}}$$. But all I got was an equally weird differential equation:

$$\frac{df}{dx}=\frac{1+\frac{1+\frac{1+\frac{{\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}}}{\frac23\frac{\left(\frac{\left(f(x)^2-x\right)^2-x}{2}\right)^2-x}{3}}}{\frac22\frac{\left(f(x)^2-x\right)^2-x}{2}}}{\frac21\left(f(x)^2-x\right)}}{2f(x)}$$

Any ideas as to how I might proceed?/Any alternative (hopefully less tedious, but regardless) methods that might work?

I created a small program to play with this. The exact answer (perhaps as an infinite series) may contain $$\sqrt{6}+1/2+...$$ somewhere in it, because as you increase the number replacing 6, the radical approaches $$\sqrt{x}+1/2$$. Of course, this term just comes from the binomial series for $$\sqrt{6+\sqrt{6}}$$.

I also got nothing on the inverse symbolic calculator.

Here's another possible approach: one may consider the sequence of polynomials:

$$P_1:x^2-6=x$$ $$P_2:\left(\frac{x^2-6}2\right)^2-6=x$$ $$P_3:\left(\frac{\left(\frac{x^2-6}2\right)^2-6}3\right)^2-6=x$$

Formed by taking recurrent approximations to the infinite radical. The limit of $$P_n$$ as $$n\to\infty$$ is the root of some function with a power series expansion that can perhaps be calculated in this form. But what is the power series expansion?

Note that the polynomial gets very complicated very quick. E.g. here's $$P_5$$:

$$\frac{x^{32}}{2751882854400}-\frac{x^{30}}{28665446400}+\frac{43x^{28}}{28665446400}-\frac{91x^{26}}{2388787200}+\frac{121x^{24}}{191102976}-\frac{53x^{22}}{7372800}+\frac{11167x^{20}}{199065600}-\frac{4817x^{18}}{16588800}+\frac{57659x^{16}}{66355200}-\frac{x^{14}}{1382400}-\frac{9491x^{12}}{1382400}+\frac{367x^{10}}{12800}-\frac{2443x^8}{46080}+\frac{179x^6}{9600}+\frac{2233x^4}{9600}-\frac{71x^2}{160}-x-\frac{33359}{6400}=0$$

• why wouly you expect anything pretty to happen ? Jul 3, 2018 at 13:00
• What is the limit value ? With estimates I get about $\,\sqrt{6}+\frac{1}{\sqrt{2}}\,$ but perhaps that's too much ? Or not enough ? (I have no programm with me to calculate it.) Jul 3, 2018 at 13:54
• Bravo for typing the expression for $df/dx$ in $\LaTeX$ though :) Jul 3, 2018 at 14:15
• @user90369 I ran a quick program to check it out (and the value it gives is right to all the given decimal places) -- unfortunately, yours goes wrong at the third decimal place. It's definitely not $\sqrt{10}$, though. Jul 3, 2018 at 16:14
• @AbhimanyuPallaviSudhir : Thanks a lot for checking it, very kind of you. Jul 3, 2018 at 17:43

Let $$G:=\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}}$$ Then define $$F:=G^2-6=\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}$$ which is easier to work with. Following on from this method, we can immediately match up $n$ and $x$. They are $n=1$ and $x=2$ (as can be observed in the radical).
Finally, we find $a$. The value of $6$ corresponds to $ax+(n+a)^2=2a+(1+a)^2$ so we solve $$6=a^2+4a+1\implies(a-1)(a+5)=0\implies a=1,-5.$$
The result is given as $$F=x+n+a=3+a$$ and since $F$ is clearly non-negative, we have that $a=1$ so $$G=\sqrt{6+F}=\sqrt{6+3+1}=\color{red}{\sqrt{10}}.$$
• This is evidently incorrect -- to use that expression, you need $ax+(n+a)^2=6$, $a(x+n)+(n+a)^2=6$, $a(x+2n)+(n+a)^2=6$, etc. -- only the first of these is true when you set $a=1$ (or any non-zero number). The infinite radical you are evaluating is $\sqrt {6 + 2\sqrt {7 + 3\sqrt {8 + 4\sqrt {9...} } } }$ ... This is why the question is tougher than Ramanujan's original problem, $a$ must equal zero for the term "6" to remain constant, but that isn't possible. Jul 3, 2018 at 16:04
• Well spotted. The value of $\sqrt{10}$ can thus be an upper bound. I'll see what I can make out of this problem this week. Jul 3, 2018 at 16:33
• TLDR; $$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6..}}}}<\sqrt{6+\sqrt{6+2\sqrt{7+3\sqrt{8+4}}}}$$ $$\sqrt{6+2\sqrt{7+3\sqrt{8+4\sqrt{..}}}}=4$$ $$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6..}}}}<\sqrt{10}$$ Jan 25, 2020 at 8:14