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

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?

• I think it can be solved using Stolz–Cesàro theorem but I really don't know how to proceed. Commented Apr 11, 2021 at 12:49
• The result depends on the initial condition, Stolz-Cesaro does not look promising here. Commented Apr 11, 2021 at 15:35
• There is a typo in the bounty text, it should be $L = \lim_{n \to \infty}(3-a_n)6^n$. Commented Apr 15, 2021 at 10:36
• Almost, but I think Somos is closer, with the reference to the Koenigs function (and Schröder's equation). Commented Apr 17, 2021 at 11:30
• 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 $f(3) = 0$. Commented Apr 18, 2021 at 20:20

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.

• 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 $a_1 = 1$. Commented Apr 16, 2021 at 20:01
• @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? Commented Apr 16, 2021 at 21:29
• Yep, that's correct. The square roots would have been annoying to deal with. Commented Apr 16, 2021 at 23:49
• @VarunVejalla OK, if you don't mind, I'll add this to my post? Commented Apr 17, 2021 at 7:41
• @VarunVejalla As I mention it is actually just the translate so that (in my notation) you expand at $u=0$ and not $w=3$. Commented Apr 17, 2021 at 22:14

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

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.

• Can you please provide a bit more detail on the origin of $F(x, q)$? The solution looks deeply mysterious!! Commented Apr 16, 2021 at 2:48
• Just curious, did Ramanujan have anything to do with this technique? Even if he didn't, he would have liked this kind of approach. Commented Apr 16, 2021 at 3:43
• @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! Commented Apr 16, 2021 at 4:00
• Perhaps I am overlooking something obvious, but how did you calculate $x_0$ from the given data? Commented Apr 16, 2021 at 18:09
• I would also love a reference for this machinery, if you have one! (+1, ofc) 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 $$(3-a_{n+1})6^{n+1} = \frac{6}{3+\sqrt{6+a_n}} (3-a_n)6^n$$

thus

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

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.

• @lonestudent Read the first sentence of his proof (the part in bold specifically)
– Anon
Commented Apr 11, 2021 at 14:57
• 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. Commented Apr 11, 2021 at 15:03
• @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 $\sqrt 6$? Commented Apr 11, 2021 at 18:41
• @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! Commented Apr 11, 2021 at 18:51
• @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. 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 $$3, then $$3 for all $$n\geq1$$
2. If $$0\leq a_0<3$$, then $$a_{n-1} for all $$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} 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} 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}

Remarks:

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 $$b_n and convergence follows immediately. This can be explained in part to the fact that $$f'(x)=\big(2\sqrt{6+x}\big)^{-1}$$ is a decreasing function, and so $$f'(y) whenever $$x<3. So convergence tens to be faster to the right of $$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 $$-6 \le a_{1} <10$$

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

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

• You switch from “approximate/asymptotic equality” ($\simeq$) to equality at various places, the equations (3) and (4) do surely not hold as written. Commented Apr 15, 2021 at 10:18
• Thanks for the hint. I thought is was clear what is meant from the context. Commented Apr 15, 2021 at 13:00
• 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 $6^n \delta_n$. 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

$$6^{-n-1}b_{n+1}+3=\sqrt{6+6^{-n}b_n+3}$$

and

$$b_{n+1}=6^{n+1}\left(\sqrt{9+6^{-n}b_n}-3\right)=\frac{6b_n}{\sqrt{9+6^{-n}b_n}+3},$$

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.

• 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? Commented Apr 11, 2021 at 16:18
• @lonestudent: I don't believe so.
– user65203
Commented Apr 11, 2021 at 16:19