Solving Recurrence Relation with Exponential Relationship I've come across a recurrence relationship in a modeling problem given by the following:
$$ a_{n+1} = a_{n} + \gamma a_{n} e^{-a_{n}^2} \qquad a_0, \gamma > 0. $$
My intuition tells me that the sequence has an accumulation point as a result of the exponential decay, however, I'm struggling to find a way to show this formally. Does anyone have any resources or means of solving this kind of relation or finding an upper bound?
 A: As in the comment, the sequence is increasing and $a_n \xrightarrow{n\to+\infty}+\infty$.
The speed of convergence is extremely low, we will find the upper bound.
It is easy to prove the two lemmas below

Lemma 1: For any $\alpha>1$, the function $x\mapsto x^\alpha e^{-x^2}$ has its maximum
$m(\alpha):=\left(\frac{\alpha}{2}\right)^\alpha
> e^{-\frac{\alpha^2}{4}}$ and for all $x> \frac{\alpha}{2}$,
$$x^\alpha e^{-x^2}<m(\alpha)$$
Lemma 2: For $\alpha>1$, The function $x \mapsto(1+x)^{\alpha}-(1+2\alpha x)$ is negative for all $0<x<2^{\frac{1}{a-1}}-1$

Return back to our sequence, as $a_n \xrightarrow{n\to+\infty}+\infty$, there exists an $N_1$ such that $a_n>\frac{\alpha}{2}$ for all $n>N_1$, applying the lemma 1, we have
$$
a_{n+1}=a_n (1+\gamma e^{-a_n^2}) <a_n(1+\gamma m(\alpha)a_n^{-\alpha})
\Longrightarrow a_{n+1}^{\alpha}<a_n^{\alpha}(1+\gamma m(\alpha)a_n^{-\alpha})^{\alpha} \tag{1}
$$
And also, $a_n^{-\alpha} \xrightarrow{n\to+\infty}0$, there exists an $N_2$ such that $a_n^{-\alpha}<2^{\frac{1}{a-1}}-1$ for all $n>N_2$, applying the lemma 2, we have
$$\begin{align}
(1)&\Longrightarrow a_{n+1}^{\alpha}<a_n^{\alpha}(1+2\gamma m(\alpha)a_n^{-\alpha})\\
&\Longrightarrow a_{n+1}^{\alpha} - a_n^{\alpha}< 2 \gamma m(\alpha)  \qquad \forall n> N:=\max\{N_1,N_2\} \\
&\Longrightarrow a_n^{\alpha}<2 \gamma m(\alpha) (n-N) + a_N^{\alpha} 
\end{align}$$
We notice that $N$ depends on $\alpha$, so, the upper bound can be
$$\color{red}{a_n < \left(2 \gamma m(\alpha) n + a_N^{\alpha} - 2N \gamma m(\alpha)\right)^{\frac{1}{\alpha}}  \qquad   \forall \alpha >1}$$
For $n\to +\infty$, the speed of convergence is less than $n^{\frac{1}{\alpha}}$, for all $\alpha >1$, which is extremely low.

Here is the sketch of proof for $\mathcal{O}(\sqrt{\ln(n)})$
We have
$$\begin{align}
&\Longleftrightarrow a_{n+1}^2 = a_n^2 (1+\gamma e^{-a_n^2})^2\\
&\Longleftrightarrow e^{-a_{n+1}^2} =e^{-a_n^2 (1+\gamma e^{-a_n^2})^2 }\\
\end{align}$$
Denote $x_n = e^{-a_n^2}$, we have $x_n \xrightarrow{n\to+\infty}0$
$$\begin{align}
&\Longleftrightarrow x_{n+1} =x_{n}^{(1+\gamma x_n)^2 } = \exp{\left((1+\gamma x_n)^2\ln(x_n)\right)}\\
&\Longleftrightarrow x_{n+1} = x_{n}\cdot\exp{\left((2\gamma +x_n)x_n\ln(x_n)\right)}\\
\end{align}$$

Lemma 3: For $x>0$ and sufficiently small, $e^{x\ln (x)}<1-x^{\alpha}\ln(x)$ for all $\alpha >0$


Lemma 4: For $x<0$ and $|x|$ is sufficiently small, $(1-x)^{-\alpha}<1-x$ for all $\alpha >0$

Applying the lemma 3 and lemma 4, given the fact that $2\gamma +x_n < 3\gamma$ for $n$ sufficiently large, we have
$$\begin{align}
&\Longrightarrow x_{n+1} < x_{n}\cdot (1 - x_n^{\alpha + 1}\ln(x_n))  \qquad \forall \alpha >0 \\
&\Longleftrightarrow x_{n+1}^{-\alpha}<x_{n}^{-\alpha}(1 - x_n^{\alpha+1}\ln(x_n))^{-\alpha}<x_{n}^{-\alpha}(1 - x_n^{\alpha+1}\ln(x_n))\\
&\Longleftrightarrow x_{n+1}^{-\alpha} - x_{n}^{-\alpha}< - x_n\ln(x_n)
\end{align}$$

Lemma 5: there exists $\beta>0$ such that $- x\ln(x) < \beta$ for $0<x<1$

Apllying the lemma 5, we can deduce that
$$\begin{align}
& x_{n}^{-\alpha} < \beta (n-N) +x_N^{-\alpha}  \qquad \forall \alpha>0\\
&\Longleftrightarrow e^{a_n^2} < (\beta (n-N) +x_N^{-\alpha})^{\frac{1}{\alpha}} \\
&\Longleftrightarrow a_n < \frac{1}{\sqrt{\alpha}}\sqrt{\ln(\beta (n-N) +x_N^{-\alpha})}  \qquad \forall \alpha>0\\
\end{align}$$
We can deduce that the upper bound is less than $\mathcal{O}(\sqrt{\ln(n)})$
