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 $P_5$:


See What is the region of convergence of $x_n=\left(\frac{x_{n-1}}{n}\right)^2-a$, where $a$ is a constant?

  • 3
    $\begingroup$ why wouly you expect anything pretty to happen ? $\endgroup$
    – mercio
    Commented Jul 3, 2018 at 13:00
  • $\begingroup$ 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.) $\endgroup$
    – user90369
    Commented Jul 3, 2018 at 13:54
  • 5
    $\begingroup$ Bravo for typing the expression for $df/dx$ in $\LaTeX$ though :) $\endgroup$
    – TheSimpliFire
    Commented Jul 3, 2018 at 14:15
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jul 3, 2018 at 16:14
  • $\begingroup$ @AbhimanyuPallaviSudhir : Thanks a lot for checking it, very kind of you. $\endgroup$
    – user90369
    Commented Jul 3, 2018 at 17:43

1 Answer 1


Currently not correct answer; but I keep it for the record (and hopefully whenever I manage to make any progress on it)

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 $$G=\sqrt{6+F}=\sqrt{6+3+1}=\color{red}{\sqrt{10}}.$$

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jul 3, 2018 at 16:04
  • $\begingroup$ 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. $\endgroup$
    – TheSimpliFire
    Commented Jul 3, 2018 at 16:33
  • $\begingroup$ TLDR; $$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6..}}}}<\sqrt{6+\sqrt{6+2\sqrt{7+3\sqrt{8+4}}}}$$ $$\sqrt{6+2\sqrt{7+3\sqrt{8+4\sqrt{..}}}}=4$$ $$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6..}}}}<\sqrt{10}$$ $\endgroup$
    – Mourad
    Commented Jan 25, 2020 at 8:14

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