Does this recurrence have a closed form limit $x_{n+1}=x_n-\frac{a}{3^{2n+1}x_n}$?

I have a first order nonlinear recurrence relation:

$$x_{n+1}=x_n-\frac{a}{3^{2n+1}x_n}$$

Here $a,x_0$ are positive constants and $a<x_0$. (Also $x_0=A+B$ and $a=(A−B)^2$, for some $A,B>0$).

For these conditions the recurrence quickly converges to a certain limit, which depends only on $a,x_0$:

$$\lim_{n \to \infty}x_n=X(a,x_0)$$

I don't know if this limit has closed form or not, and there is no general method for dealing with nonlinear recurrence relations.

Can $X(a,x_0)$ have a closed form and how to obtain it?

I don't need the explicit expression for $x_n$, only the limit.

I tried to turn it into a differential equation, but I don't know if I've done it correctly, and how the solution to the ODE relates to the original problem:

$$x_{n+1}-x_n=-\frac{a}{3^{2n+1}x_n}$$

$$\frac{df(t)}{dt}=-\frac{b}{3^{2t} f(t)}$$

$$\frac{1}{2} f^2=\frac{b}{2\ln 3} 3^{-2t}+C$$

$$f(t)=\sqrt{\frac{a}{3\ln 3} 3^{-2t}+C}$$

If I set:

$$x_n=\sqrt{\frac{a}{3\ln 3} 3^{-2n}+C}$$

I get:

$$C=x_0^2-\frac{a}{3\ln 3}$$

$$\lim_{n \to \infty}x_n=\sqrt{C}=\sqrt{x_0^2-\frac{a}{3\ln 3}}$$

But that's not correct. Does this work only with linear recurrences?

I have also inverted the recurrence:

$$x_n=\frac{1}{2} \left(x_{n+1}+\sqrt{x_{n+1}^2+\frac{4a}{3^{2n+1}}} \right)$$

This works correctly, but I'm not sure how it may help.

I would be astounded if the limit has a nice closed form. Here are some facts I have investigated.

Let $f(z) = 3z - \frac{1}{z}$. Then we easily check that $y_n = \frac{3^n}{\sqrt{a}} x_n$ satisfies $y_{n+1} = f(y_n)$. Thus $x_n$ has the following general form

$$ x_n = \frac{\sqrt{a}}{3^n} f^{\circ n}(y_0) = \sqrt{a} F_n\left(\frac{x_0}{\sqrt{a}}\right), \qquad F_n(z) := z \prod_{k=0}^{n-1} \left(1 - \frac{1}{3f^{\circ k}(z)^2} \right). $$

So it suffices to consider the case where $a = 1$ and then investigate the limit of $F_n(z)$ if exists. To this end, we claim the following:

Proposition. Let $I = [-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$. Then the limit $F(z) = \lim_{n\to\infty} F_n(z)$ converges and defines a holomorphic function on $\Bbb{C}\setminus I$. Moreover,

1. There exists a finite Borel measure $\mu$ on $\Bbb{R}$ which is supported on $I$ and satisfies \begin{align*} F(z) &= z - \int_I \frac{\mu(d\lambda)}{z-\lambda} \\ &= z - \frac{3}{8z} - \frac{9}{640z^3} - \frac{2241}{465920z^5} - \cdots \quad \text{as } z \to \infty. \end{align*}

2. $F(f(z)) = 3F(z)$.

The proof of this fact is at the end of this answer. Here are some comments:

• Identifying $F$ amounts to identifying the measure $\mu$. Although not entirely sure, it seems to me that $\mu$ is discrete and $\operatorname{supp}(\mu)$ is a Cantor-like set. That is, poles aggregate at uncountably many points and they do not average out. I can hardly imagine of an elementary function which exhibits this kind of behavior, so I suspect that $F$ is not an elementary function.

• On $(\frac{1}{\sqrt{2}}, \infty)$ the function $f$ has inverse. If we denote by $g = f^{-1}$ the inverse, then for each $\epsilon > 0$ we can prove that

  $$ g^{\circ n}\left(\frac{1}{\sqrt{2}} + \epsilon\right) = \frac{1}{\sqrt{2}} + \frac{\ell + o(1)}{5^n} \qquad \text{as } n\to\infty \tag{*}$$

  for some constant $\ell = \ell(\epsilon) \geq$ which depends on $\epsilon$. Then in view of the identity

  $$ F\left(\frac{1}{\sqrt{2}} + \epsilon\right) = 3^n F\left(g^{\circ n}\left(\frac{1}{\sqrt{2}} + \epsilon\right) \right), $$

  if $F(\frac{1}{\sqrt{2}} + \epsilon) \sim c \epsilon^{\alpha}$ for some constants $c > 0$ and $\alpha$, then we must have $\alpha = \frac{\log 3}{\log 5}$. Although I have not formally written down my idea, using a detailed version of $\text{(*)}$ I checked that $F(\frac{1}{\sqrt{2}} + \epsilon) = \Theta(\epsilon^{\alpha})$ for this $\alpha$, where $\Theta$ is a Landau asymptotic notation.

Proof of Proposition. By the extra assumption $a =1$, we have $x_n = F_n(x_0)$ and $y_n = f^{\circ n}(x_0)$. From this, we have the following relations.

$$ F_{n+1}(z) = F_n(z) - \frac{1}{3^{2n+1}F_n(z)}, \qquad F_n(z) = 3^{-n} f^{\circ n}(z). \tag{1}$$

Using the first relation in $\text{(1)}$, we have

$$ \operatorname{Im}(F_{n+1}(z)) = \operatorname{Im}(F_n(z))\left( 1 + \frac{1}{3^{2n+1}|F_n(z)|^2} \right). $$

In particular, $F_n$ is a Nevanlinna function and thus we can write write $F_n(z)$ as

$$F_n(z) = z - \int_{\Bbb{R}} \frac{\mu_n(d\lambda)}{z-\lambda} $$

for some point mass $\mu_n$ on $\Bbb{R}$. (This can also be proved directly by some tedious algebra.) This representation shows that $F_n(z) = z - \mu_n(\Bbb{R})z^{-1} + \mathcal{O}(z^{-2})$ as $z \to \infty$. Plugging this to the recurrence relation in $\text{(1)}$, we find that

$$\mu_{n+1}(\Bbb{R}) = \mu_n(\Bbb{R}) + \frac{1}{3^{2n+1}}. $$

In particular, the total mass of $(\mu_n)$ is bounded. On the other hand, if $x \in \Bbb{R}\setminus I$, then we easily check that $|f(x)| > |x|$. This implies that

$\text{(2)}$ $F_n$ is finite on $\Bbb{R}\setminus I$, and thus $\mu_n$ is supported on $I$.

$\text{(3)}$ $F_n$ converges on $\Bbb{R}\setminus I$, since $F_n(x)$ is monotone and bounded for each $x \in \Bbb{R}\setminus I$.

So it follows from $\text{(2)}$ that $(\mu_n)$ is weakly compact and hence $F_n$ is a normal family on $\Bbb{C}\setminus I$. Then the first claim follows from the identity theorem together with $\text{(3)}$.

Once we know that $F_n$ converges on $\Bbb{C}\setminus I$, then the second claim follows by taking the limit to $F_n(f(z)) = 3F_{n+1}(z)$. Using this, we can compute the Laurent expansion of $F$ near $\infty$. Since $F$ is an odd function, with $a_n = \int_{\Bbb{R}} \lambda^{2n} \, \mu(d\lambda)$ we have

$$ F(z) = z - \sum_{n=0} \frac{a_n}{z^{2n+1}}. $$

Plugging this to the identity $F(f(z)) = 3F(z)$ produces an system of equations for $(a_n)$, from which we can compute $(a_n)$ at least theoretically.