Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$.

First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the form $0 \cdot \infty$.

Moreover, we obtained a similar result here, which is related to the form $\lim\limits_{n \to \infty} (a_n-3)9^n$.

How should I proceed?

  • 2
    $\begingroup$ I think it can be solved using Stolz–Cesàro theorem but I really don't know how to proceed. $\endgroup$
    – DaifM
    Commented Apr 11, 2021 at 12:49
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    $\begingroup$ The result depends on the initial condition, Stolz-Cesaro does not look promising here. $\endgroup$
    – leonbloy
    Commented Apr 11, 2021 at 15:35
  • $\begingroup$ There is a typo in the bounty text, it should be $L = \lim_{n \to \infty}(3-a_n)6^n$. $\endgroup$
    – Martin R
    Commented Apr 15, 2021 at 10:36
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    $\begingroup$ Almost, but I think Somos is closer, with the reference to the Koenigs function (and Schröder's equation). $\endgroup$
    – rtybase
    Commented Apr 17, 2021 at 11:30
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    $\begingroup$ If you let the limit be a function of $a_0$ as $f(a_0)$, then $f'(a_0) = \lim_{n \to \infty} \frac{3^n}{a_1\cdot a_2\cdots a_n}$, where $a_1, a_2, \cdots, a_n$ are all functions of $a_0$ and $f(3) = 0$. $\endgroup$ Commented Apr 18, 2021 at 20:20

7 Answers 7


The function, $a_1 \mapsto \Phi(a_1) = \lim_{n\to \infty} 6^n (a_n-3)$ is a well-known object in complex dynamics and known as a linearization map. It solves a conjugation problem, which was first introduced by Ernst Schröder in 1871 to study iterations of rational functions. You may find an account e.g. in Carleson and Gamelin: Complex Dynamics. In chap II.2 we have the following:

Theorem 2.1: Suppose an analytic function $f$ has an attractive fixed point $p$ with multiplier $\lambda=f'(p)$ satisfying $0<|\lambda|<1$. Then there is a conformal map $\zeta=\phi(z)$ (unique up to scaling) of a neighborhood of $p$ onto a neighborhood of $0$ which conjugates $f(z)$ to the linear function $g(\zeta)=\lambda \zeta$.

The proof goes through the construction of the limit you mention. To be more explicit and using your example: Let $D={\Bbb C}\setminus (-\infty;-6]$ be the slit complex plane. Then $f(z)=\sqrt{z+6}$ defines a conformal map of $D$ into the right half plane ${\Bbb H}_+\subset D$. Thus, for all $z\in D$: $$|f(z)-3| = \left|\frac{z-3}{\sqrt{z+6}+3}\right| \leq \frac13 |z-3|$$ Thus we have convergence of $f^n(z)$ to the fixed point $p=3$, uniformly on compact subsets of $D$. We also have $\lambda=f'(3) = \frac16$.

The proof of the theorem is then to show that the sequence of maps ($n\geq 0$): $$ \phi_n(z) = \lambda^{-n} (f^n(z)-p) = 6^n (f^n(z)-3)$$ converges uniformly on a neighborhood of $p$. Pick $z_0=z\in D$ and let $z_n=f^n(z_0)$ (I prefer starting at index 0 so that we iterate $f$ as many times as the linear map) we have: $$ \phi_{n+1}(z) = 6^n (z_n-3) \frac{6}{\sqrt{z_n+6}+3} = \phi_n(z) \left(1 + \frac{3-\sqrt{z_n+6}}{\sqrt{z_n+6}+3}\right) = \phi_n(z) \left( 1 + \epsilon(z_n)\right) $$ where $$\epsilon(z_n) = \frac{3-z_n}{\left( \sqrt{6+z_n} + 3\right)^2}. $$ One has $|\epsilon(z_n)| \leq \frac19 |z_n-3| \leq \frac{1}{3^{n+2}} |z-3|$ thus going to zero exponentially fast and uniformly on compact subsets in $D$. It follows that the limit $$\phi(z) = \lim_n \phi_n(z)=(z-3)\prod_{n\geq 1} (1+\epsilon(z_n))$$ exists and is holomorphic in $D$. The function verifies $\phi'(3)=1$ whereas your function has an extra factor $6$ since your index starts with $n=1$ rather than zero. So your function is really $\Phi(a_1)=6\phi(a_1)$. The formula is numerically precise and efficient: $6\phi(\sqrt{6})=-3.36565753974384...$ in agreement with other results posted.

To see that $\phi$ actually yields a solution to the Theorem, note that by construction we have $\lambda^{-1} \phi_n\circ f = \phi_{n+1}$. By uniform convergence we may take limits to get $\lambda^{-1} \phi\circ f = \phi$ or $\phi \circ f = \lambda \phi = g\circ \phi$, with $g(\zeta)=\lambda \zeta$ being the linear map.

A Taylor expansion of $\phi$ may be obtained from the functional equation $6\phi(\sqrt{z+6}) = \phi(z)$ by expanding at $z_0=3$ on both sides and identifying coefficients. A far more elegant way to do the computations was suggested in comments by @VarunVejalla: Solving $w=\sqrt{z+6}$ gives $z=w^2-6$, i.e. the relation $6\phi(w)= \phi(w^2-6)$. To simplify notation set $w=3+u$. Then $\psi(u)=\phi(3+u)$ is to be expanded around 0 and verifies $6\psi(u)=\phi((3+u)^2-6) = \psi(6u+u^2)$. Inserting $\psi(u)=u + \sum_{n\geq 2} a_n u^n$ yields the identity between Taylor series:

$$ 6u + \sum_{n\geq 2} 6 a_n u^n = (6u+u^2) + \sum_{n\geq 2} a_n (6u +u^2)^n $$ or $$ \sum_{n\geq 2} a_n ((6u+u^2)^n-6u^n) = -u^2 $$ Expanding the binomial and identifying coefficients you may solve recursively to get a formula for $a_n$ which only depends upon the $a_k$ with $n/2 \leq k \leq n$. You get $a_1=1$ and then $$ a_n = \frac{1}{6-6^n} \sum_{k =\lceil n/2 \rceil}^{n-1} \pmatrix {k \\ n-k} 6^{2k-n} a_k $$ The first terms: $$ \psi_4(u) = u -\frac{1}{30}u^2 + \frac{1}{525} u^3 - \frac{181}{1354500} u^4. $$ With 12 terms one gets $6 \psi_{12}(\sqrt{6}-3) = 3.365657539743842... $ correct to 14 digits.

  • $\begingroup$ To add on to this, the coefficients of the Taylor series of $\phi(z)$ can be written somewhat explicitly as $a_n=\frac{1}{6-6^n}\sum_{k=\lceil n/2 \rceil}^{n-1}\binom{k}{n-k}a_k 6^{2k-n}$ for $n \ge 2$, with $a_0 = 0$ and $a_1 = 1$. $\endgroup$ Commented Apr 16, 2021 at 20:01
  • $\begingroup$ @VarunVejalla Thanks for the remark. And nice formula. I suppose that you develloped the version $6\psi(u)=\psi(6u+u^2)$ to obtain this? $\endgroup$
    – H. H. Rugh
    Commented Apr 16, 2021 at 21:29
  • $\begingroup$ Yep, that's correct. The square roots would have been annoying to deal with. $\endgroup$ Commented Apr 16, 2021 at 23:49
  • $\begingroup$ @VarunVejalla OK, if you don't mind, I'll add this to my post? $\endgroup$
    – H. H. Rugh
    Commented Apr 17, 2021 at 7:41
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    $\begingroup$ @VarunVejalla As I mention it is actually just the translate so that (in my notation) you expand at $u=0$ and not $w=3$. $\endgroup$
    – H. H. Rugh
    Commented Apr 17, 2021 at 22:14

The following solution is a special case of a general method.

The question asks

Let $a_ 1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim_{n \to \infty} (a_n-3)6^n$.

With $\,q\,$ as a parameter, define the function $$ F(x) := \sum_{n=0}^\infty c_n \frac{x^n}{f_n}\;\; \text{ where }\;\; f_n:= \prod_{k=1}^n (1-q^k) \tag{1} $$ and where $\,|q|\ne1.\,$ Define the constants $$ L := q/2, \quad \text{ and } \quad K := L^2-L. \tag{2} $$ Also define $$ a_n := A\left(\frac{x_0}{q^n}\right) \;\; \text{ where } \;\; A(x) := L - q\, x\, F(x) \tag{3} $$ and where $\,x_0 = A^{-1}(a_0)\,$ depends only on $\,q\,$ and $\,a_0.\,$

Tthe value of $\,x_0\,$ computed this way does not suffer from floating point rounding problems.

The equation $$ a_{n+1} = \sqrt{K+a_n} \quad \text{ or } \quad a_{n+1}^2 = K+a_n \tag{4} $$ implies that the coefficients of $\,F(x)\,$ as polynomials in $\,q\,$ satisfy $$ c_0 = 1, \quad c_{n+1} = \sum_{k=0}^n c_{n-k}\,c_k\, \frac{f_n}{f_k f_{n-k}}. \tag{5} $$

Note that $$ F(0) = 1,\; A(0) = L \; \text{ and }\; a_n\to L. \tag{6} $$ The approach of $\,a_n\,$ to the limit $\,L\,$ is given by $$ L - a_n = q \, \frac{x_0}{q^n} F\left(\frac{x_0}{q^n} \right) \approx q\frac{x_0}{q^n}. \tag{7} $$ This implies $$ \lim_{n\to\infty}(a_n-L)\,q^n = -q\,x_0. \tag{8} $$

In the case in the question, $$ q = 6,\; L = 3,\; a_0 = 0,\; x_0 \approx 0.56094292329064. \tag{9} $$

In the case from Art of Problem Solving Online, $$ q = 4,\; L = 2,\; a_0 = 0,\; x_0 = \frac{\pi^2}{16},\; F(x) = \frac{\sin(\sqrt{x})^2}x. \tag{10} $$

NOTE: About my method. Some of it is based on Koenigs function but my version is more constructive. If we have a function $\,T(x)\,$ and define a sequence by $\,a_{n+1} = T(a_n)\,$ where $\,a_n\to 0\,$ such that $\, a_n \approx c/q^n,\,$ then use the Ansatz $\, a_n = F(x/q^n)\,$ for some function $\,F\,$ with a power series expansion in $\,x\,$ with coefficients that depend on $\,q.\,$ The coefficients are uniquely determined by the function $\,T.$ The mode of convergence determines the proper Ansatz. For example, $\,T(x) := x-x^2\,$ requires a different Ansatz.

NOTE: If $\,q=1\,$ then the convergence $\,a_n\to \frac14\,$ is much slower and the above analysis does not hold. Instead the one for $\,T(x) := x-x^2\,$ is needed. Something similar if $\,q=-1\,$ and other roots of unity.

NOTE: I was going to mention chapter 8.3 of Asymptotic Methods in Analysis by de Bruijn which I am familiar with but didn't have the exact page number.

  • 3
    $\begingroup$ Can you please provide a bit more detail on the origin of $F(x, q) $? The solution looks deeply mysterious!! $\endgroup$
    – Paramanand Singh
    Commented Apr 16, 2021 at 2:48
  • $\begingroup$ Just curious, did Ramanujan have anything to do with this technique? Even if he didn't, he would have liked this kind of approach. $\endgroup$
    – Paramanand Singh
    Commented Apr 16, 2021 at 3:43
  • $\begingroup$ @Somos Please let me ask...How did you invent all this? I would be glad if you want to say / share. How did you invent it all? Does your answer prove that the value of the limit cannot be expressed in $\pi$ ? And also, does your answer prove that the limit cannot be expressed by any "known" constant in mathematics?.. Thank you very much! $\endgroup$ Commented Apr 16, 2021 at 4:00
  • 1
    $\begingroup$ Perhaps I am overlooking something obvious, but how did you calculate $x_0$ from the given data? $\endgroup$
    – Martin R
    Commented Apr 16, 2021 at 18:09
  • 1
    $\begingroup$ I would also love a reference for this machinery, if you have one! (+1, ofc) $\endgroup$ Commented Apr 16, 2021 at 19:45

Here's a proof that $b_n:=(3-a_n)6^n$ converges to a finite limit. This is merely a proof of existence of a finite limit (which I don't think is trivial).

It is shown here that $a_n$ is strictly increasing and has limit $3$.

Note that $$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}$$

hence $$(3-a_{n+1})6^{n+1} = \frac{6}{3+\sqrt{6+a_n}} (3-a_n)6^n$$


$$\frac{b_{n+1}}{b_n} = \frac{6}{3+\sqrt{6+a_n}}>1$$

Let us show that $\sum_n \ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)$ converges. From $$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}\leq \frac 13 (3-a_n)$$ one readily finds that $3-a_n = O(\frac 1{3^n})$, hence $a_n = 3 + O(\frac 1{3^n})$, thus $$\ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big) = \ln\Big(\frac{6}{3+\sqrt{9+O(\frac 1{3^n})}}\Big) = \ln(1+O(\frac 1{3^n})) = O(\frac 1{3^n})$$ Hence$\sum_n \ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)$ converges.

Thus $\sum_n \log(\frac{b_{n+1}}{b_n})$ converges, hence the sequence $\log b_n$ converges to a finite limit, hence $b_n$ converges as well to some positive real.

Therefore, the sequence $(a_n-3)6^n$ converges to a negative real number.

  • 2
    $\begingroup$ @lonestudent Read the first sentence of his proof (the part in bold specifically) $\endgroup$
    – Anon
    Commented Apr 11, 2021 at 14:57
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    $\begingroup$ Numerical experiments (with PARI/GP) indicate that the limit of $(3-a_n)6^n$ is approximately $3.36565753974384094$. It may well be that this cannot be expressed in terms of elementary functions, the inverse symbolic calculator does not recognize that number. – Proving that the limit exists at all is a substantial contribution in my opinion. $\endgroup$
    – Martin R
    Commented Apr 11, 2021 at 15:03
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    $\begingroup$ @Martin $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=2\cos \frac{\pi}{2^{n+1}}$$ Is it not possible to construct a similar formula for $\sqrt 6$? $\endgroup$ Commented Apr 11, 2021 at 18:41
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    $\begingroup$ @Martin I actually kept the comment short:. I knew you knew this formula for sure :) Please let me edit my comment. Why can't the formula finding method used for $\sqrt 2$ work for $\sqrt 6$? so what makes the $\sqrt 2$ special? I don't know how to find this formula. I only know the proof by induction. in short: does the formula finding method for $\sqrt 2$ work only in "special cases"? Thank you for your explanation! $\endgroup$ Commented Apr 11, 2021 at 18:51
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    $\begingroup$ @lonestudent: The factor $2^{n+1}$ in your example comes from the “half-angle formula” for the cosine. Here we would need a different factor, but there are no equally “simple” formulas for $\cos(x/c)$ for $c \ne 2$, at least none that I know of. I am not saying that it is not possible, just that I don't yet know how. $\endgroup$
    – Martin R
    Commented Apr 11, 2021 at 19:02

In this answer we consider the sequence $b_n=|3-a_n|6^n$, where $a_{n+1}=\sqrt{6+a_n}$ with initiial condition $a_0\geq0$. We show that $b_n$ is

  1. bounded and monotone increasing if $0\leq a_0<3$ (see \eqref{three}).
  2. bounded and monotone decreasing when $a_0>3$ (see \eqref{twop}).

As a consequence,

  1. $b_n(a_0)$ (dependence of initial condition) is convergent for all $a_0\geq0$: $$b_\infty(a_0)=|3-a_0|\prod^\infty_{n=1}\Big(1-\frac{|3-a_n|}{6}\Big)^{-1}$$
  2. When $0\leq a_0<3$, $(a_n-3)6^n=-b_n$ converges to a negative number; when $a_0>3$, $(a_n-3)6^n=b_n$ converges to a positive number.

I don't think the limit $b_\infty(a_0)=\lim_nb_n(a_0)$ can be expressed in terms of elementary functions. A numerical estimate (with double precision arithmetic) gives $b_\infty(0)\approx3.3657$.

The map $f:[0,\infty)\rightarrow[0,\infty)$ defined by $$ f(x)=\sqrt{6+x}$$ is a contraction: First notice that $f(x)\geq\sqrt{6}$ for all $x\in[0,\infty$, and \begin{align} f(x)-f(y)=\frac{y-x}{\sqrt{6+x}+\sqrt{6+y}}\tag{1}\label{one} \end{align} Then, $$|f(x)-f(y)|\leq\frac{1}{2\sqrt{6}}|x-y| $$ and so, for any $a_0\geq0$, $a_{n+1}=f(a_n)$ converges to a fixed point $x=f(x)$ with $x\geq0$, which in this case is $x=3$.

Further analysis of the map $f$ shows that

  1. If $3<a_0$, then $3<a_n<a_{n-1}$ for all $n\geq1$
  2. If $0\leq a_0<3$, then $a_{n-1}<a_n<3$ for all $n\geq1$.

We now study the speed of convergence of the sequence $a_n$. Let $d_n=|3-a_n|=|f(3)-f(a_{n-1})|$. It follows from \eqref{one} that $$d_n=\frac{d_{n-1}}{3+a_n}$$ If $0\leq a_0<3$, then \begin{align} \frac{1}{6}d_{n-1}<d_n=\frac{d_{n-1}}{3+a_n}=\frac{d_{n-1}}{6-d_n}\tag{2}\label{two} \end{align} As $a_n\xrightarrow{n\rightarrow\infty}3$, we have that $\frac{d_{n+1}}{d_n}\xrightarrow{n\rightarrow\infty}\frac{1}{6}$; consequently $\sqrt[n]{d_n}\xrightarrow{n\rightarrow\infty}\frac{1}{6}$. This alone however, does not provide convergence of $b_n=(6\sqrt[n]{d_n})^n$.

Now we prove that the sequence $b_n$ indeed converges. From \eqref{two} we have that
\begin{align} b_{n-1}=6^{n-1}d_{n-1}< 6^nd_n=b_n=\frac{6^{n-1}d_{n-1}}{1-\tfrac{d_n}{6}}=\frac{b_{n-1}}{1-\tfrac{d_n}{6}}\tag{3}\label{three} \end{align} and \begin{align} \frac{d_{n-1}}{6}<d_n\leq\frac{d_{n-1}}{3+a_0}\leq \frac{d_0}{(3+a_0)^{n-1}}\tag{4}\label{four} \end{align} Putting this together, we obtain \begin{align} b_n=b_0\prod^n_{k=1}\frac{b_k}{b_{k-1}}=b_0\prod^n_{k=1}\Big(1-\tfrac{d_k}{6}\Big)^{-1}\tag{5}\label{five} \end{align} From \eqref{four} we have that $\sum^\infty_{k=1}d_k<\infty$; from $$\prod^n_{k=1}\Big(1-\tfrac{d_k}{6}\Big)\leq \exp\Big(-\frac{1}{6}\sum^n_{k=1}d_k\Big)$$ we conclude that $b_n$ converges, and that \begin{align} b_n\xrightarrow{n\rightarrow\infty} b_0\prod^\infty_{k=1}\big(1-\tfrac{d_k}{6}\big)^{-1}\tag{6}\label{6} \end{align}


  1. An interesting observation is that when $a_0>3$, convergence of $b_n$ is much easier to check, for in this case, the inequalities in \eqref{two} are reverse, that is \begin{align} \frac{d_{n-1}}{6+d_n}=\frac{d_{n-1}}{3+a_n}=d_n<\frac{1}{6}d_{n-1}\tag{2'}\label{twop} \end{align} This implies $ b_n<b_{n-1}$ and convergence follows immediately. This can be explained in part to the fact that $f'(x)=\big(2\sqrt{6+x}\big)^{-1}$ is a decreasing function, and so $f'(y)<f'(x)$ whenever $x<3<y$. So convergence tens to be faster to the right of $3$.
  2. Notice that the limit of the sequence $b_n$ depends on the initial condition $a_0$. For $a_0=0$, simple numerical implementation gives $\lim_nb_n\approx 3.3657$; for $a_0=6$, $\lim_nb_n\approx 2.7426$

Ths approach shows that the results in question emerge quite naturally from the standard fixed-points-stability analysis of the evolution equation. Also the dependence of the limit of $b_n = (a_n -3)6^n$ from the inital value $a_1$ is discussed.

Following the common procedure to analyse the evolution equation

$$a_{n+1} = \sqrt{6+a_{n}}\tag{1}$$

we fist look for fixed points. These follow from the equation $0 = a^2-a-6$ to be $a = 3$ and $a=-2$.

Ruling out the negative solution we study the stabilty letting

$$a_{n} = 3 +\delta_{n}\tag{2}$$

where $|\delta_{n}| << 3$.

Inserting into $(1)$ gives

$$\begin{align} & 3 +\delta_{n+1} = \sqrt{6+3+\delta_{n}}=\sqrt{9+\delta_{n}}\\&=3\left( \sqrt{1+\frac{\delta_{n}}{9}}\right)\simeq 3\left( 1+\frac{1}{2}\frac{\delta_{n}}{9}\right)=3 +\frac{1}{6} \delta_{n}\end{align} $$

whence follows

$$\delta_{n+1} \simeq \frac{1}{6} \delta_{n}\tag{3}$$

The solution $\delta_{n}=\frac{c}{6^n}$ of $(3)$ with some constant $c$ shows that the fixed point $a_{\infty}=3$ is stable and from $(2)$ we find

$$a_{n} \simeq 3 + \frac{c}{6^n}\tag{4}$$

This in turn means that

$$b_{n} := \left(a_{n} - 3 \right)6^n \simeq c\tag{5}$$

So that the limit in question is the constant $c$. For $0 \lt a_{1} \lt 3$ the constant $c$ is negative, for $a_{1} \gt 3$ it is positive. In the actual case we have $a_{1} = \sqrt{6} \lt 3$, and the limit is numerically

$$b_{\infty} \simeq -3.36571\tag{6}$$.

Limit as a function of the initial value

Here is a plot showing the dependence of the limit $b_{\infty}$ on the initial value $a_{1}$ for $-6 \le a_{1} <10$

enter image description here

Some special values of the limit are: $b_{\infty}(-6) = -121.164$, $b_{\infty}(0) = -20.1939$, and $b_{\infty}(3) = 0$

The limit function is well approximated by

$$b_{\infty}(a_1)=45.1116 (a_1+6)^{0.44966}-121.164\tag{7}$$

  • 1
    $\begingroup$ You switch from “approximate/asymptotic equality” ($\simeq$) to equality at various places, the equations (3) and (4) do surely not hold as written. $\endgroup$
    – Martin R
    Commented Apr 15, 2021 at 10:18
  • $\begingroup$ Thanks for the hint. I thought is was clear what is meant from the context. $\endgroup$ Commented Apr 15, 2021 at 13:00
  • 1
    $\begingroup$ Your solution is still a bit unclear to me. I assume that (3) means $\lim_{n\to \infty} \delta_{n+1}/\delta_{n} = 1/6$. But that alone does not imply the convergence of $6^n \delta_n$. $\endgroup$
    – Martin R
    Commented Apr 15, 2021 at 14:28

Found a possible numerical approach compatible with Gabriel Romon above. Calculate $a_n$ for $n=0,1,2,...,10$.

Define $q_n=(3-a_n)6^{n}$.

Then, $\ln(1-\frac{q_n}{q_{n+1}})$ can be fitted with a straight line as a function of $n$ having $r^2=0.9983$.

If the argument of the natural logarithm here is taken to be the ratio of the first derivative to the function itself, this implies $q(n)$ is a double exponential.

To Wit:

$$q_n=(3-a_n)6^n \approx q(n)= c_1e^{(\frac{e^b}{m}e^{mn})}$$

where m and b are the slope and intercept respectively for the aforementioned line, and $c_1$ is a constant.

In my modeling, m=-1.8368, b = -0.4174.

This suggest $c_1$ is the limit in question and fitting $q(n)$ against the $q_n$ will yield $c_1$, approximating the desired result, however so far my attempts give a value of about 4.27.


With $b_n:=(a_n-3)6^{n}$ we rewrite




with $b_1=6(\sqrt 6-3)$. This proves that if $b_n$ converges, it converges to a negative value. As the denominator is asymptotic to $a+6^{-n}c$, it should be possible to find an upper bound and prove convergence.

  • $\begingroup$ Is it possible to find a general formula for $a_n?$ I think on it. Is it possible to do this using only algebra? Do you think it is worth trying to looking for a general formula? $\endgroup$ Commented Apr 11, 2021 at 16:18
  • 2
    $\begingroup$ @lonestudent: I don't believe so. $\endgroup$
    – user65203
    Commented Apr 11, 2021 at 16:19

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