We consider only a subset of the real convergent sequences spawned by the recurrence relation.
$$a_{n+1}=a_n-ba_n^{3/2}c^{n/2}$$
$$b>0 \quad\quad 0<c<1 \quad\quad 0<a_0<b^{-2}$$
There is numerical evidence that $\,a_n\,$ converges for some combinations with $\,b<0\,$ while diverges for others. I have not yet determined the convergence boundary when $\,b<0$, so only $\,b>0\,$ is considered.
We prove $\,a_n\in\mathbb{R} \,\,\,\land\,\,\, 0<a_n<b^{-2}\,$ by induction. The base case is given.
$$a_n>0 \implies a_{n+1}\in\mathbb{R}\,\,\,\land\,\,\,a_{n+1}<a_n$$
$$a_{n+1}<a_n<b^{-2} \implies a_{n+1}<b^{-2}$$
$$a_{n+1}>0 \iff a_n-ba_n^{3/2}c^{n/2}>0 \iff a_nc^n<b^{-2} \Longleftarrow a_n<b^{-2}$$
$\,a_n\,$ is a strictly decreasing sequence with a lower bound of $\,0$, so it must converge to $\,a_\infty\in\left[0,a_0\right)$.
While I do not use the following substitution in my answer, others may find it helpful:
$$a_{n+1}=a_n-ba_n^{3/2}c^{n/2} \implies a_{n+1}c^{n+1}=a_nc^nc\left(1-b\left(a_nc^n\right)^{1/2}\right)$$
$$d_n=a_nc^n \quad\quad d_{n+1}=cd_n-bcd_n^{3/2} \quad\quad d_0=a_0 \quad\quad d_\infty=0 \quad\quad d_{n+1}<d_n$$
Scratch Work:
We assume $\,\exists f:[0, 1]\to\mathbb{R}\,$ such that $\,f\,$ is right-differentiable at $\,0\,$ and the following is satisfied:
$$f\left(x\sqrt{c}\right)=f(x)-bxf^{3/2}(x)$$
$$f(1)=a_0$$
We will not attempt to prove the existence of such a function. We use induction to show $\,a_n=f\left(c^{n/2}\right)$. The base case is given.
$$f\left(c^{(n+1)/2}\right)=f\left(c^{n/2}\sqrt{c}\right)=f\left(c^{n/2}\right)-bc^{n/2}f^{3/2}\left(c^{n/2}\right)=a_n-bc^{n/2}a_n^{3/2}=a_{n+1}$$
We show $\,f(0)=a_\infty$. Since $\,f\,$ is right-differentiable at $\,0\,$, it is also right-continuous at $\,0$.
$$a_\infty=\lim_{n\to\infty}f\left(c^{n/2}\right)=f\left(\lim_{n\to\infty}c^{n/2}\right)=f(0)$$
Now we find the right derivative of $\,f\,$ at $\,0$ in two different ways.
$$\sqrt{c}f'\left(x\sqrt{c}\right)=f'(x)-bf^{3/2}(x)-\frac{3}{2}bxf'(x)f^{1/2}(x)$$
$$\sqrt{c}f'(0)=f'(0)-bf^{3/2}(0)$$
$$f'(0)=\frac{bf^{3/2}(0)}{1-\sqrt{c}}=\frac{ba_\infty^{3/2}}{1-\sqrt{c}}$$
$$f'(0)=\lim_{x\to0^+}\frac{f(x)-f(0)}{x}=\lim_{n\to\infty}\frac{f\left(c^{n/2}\right)-f(0)}{c^{n/2}}=\lim_{n\to\infty}\frac{a_n-a_\infty}{c^{n/2}}$$
$$\lim_{n\to\infty}\frac{a_n-a_\infty}{c^{n/2}}=\frac{ba_\infty^{3/2}}{1-\sqrt{c}}$$
I have numerically verified the previous limit in a few cases. Due to our assumption of the existence of a magic function, we cannot trust the above result yet. We use it as guidance for what to rigorously prove.
Rigorous Approach:
We can represent $\,a_\infty\,$ as a telescoping sum:
$$\forall{n} \quad a_\infty=a_n+\sum_{k=n}^\infty{\left(a_{k+1}-a_k\right)}$$
$$a_n-a_\infty=\sum_{k=n}^\infty{\left(a_k-a_{k+1}\right)}=\sum_{k=n}^\infty{ba_k^{3/2}c^{k/2}}$$
Since $\,a_k\,$ is a strictly decreasing sequence,
$$\sum_{k=n}^\infty{ba_\infty^{3/2}c^{k/2}}<\sum_{k=n}^\infty{ba_k^{3/2}c^{k/2}}<\sum_{k=n}^\infty{ba_n^{3/2}c^{k/2}}$$
The bounds are geometric series.
$$\sum_{k=n}^\infty{ba_\infty^{3/2}c^{k/2}}=ba_\infty^{3/2}c^{n/2}\sum_{k=0}^\infty{c^{k/2}}=\frac{ba_\infty^{3/2}c^{n/2}}{1-\sqrt{c}}$$
$$\sum_{k=n}^\infty{ba_n^{3/2}c^{k/2}}=\frac{ba_n^{3/2}c^{n/2}}{1-\sqrt{c}}$$
$$\frac{ba_\infty^{3/2}c^{n/2}}{1-\sqrt{c}}<a_n-a_\infty<\frac{ba_n^{3/2}c^{n/2}}{1-\sqrt{c}}$$
$$\frac{ba_\infty^{3/2}}{1-\sqrt{c}}<\frac{a_n-a_\infty}{c^{n/2}}<\frac{ba_n^{3/2}}{1-\sqrt{c}}$$
We use the squeeze theorem to conclude:
$$\lim_{n\to\infty}\frac{a_n-a_\infty}{c^{n/2}}=\frac{ba_\infty^{3/2}}{1-\sqrt{c}}$$
We prove $\,a_\infty\neq0$. Assume $\,a_\infty=0\,$ for contradiction. The above limit gives us something easier to contradict:
$$\lim_{n\to\infty}\frac{a_n}{c^{n/2}}=0$$
$$s_n=\frac{a_n}{c^{n/2}} \quad\quad s_0=a_0 \quad\quad s_\infty=0 \quad\quad s_n>0$$
$$s_{n+1}c^{(n+1)/2}=s_nc^{n/2}-bs_n^{3/2}c^{3n/4}c^{n/2}$$
$$s_{n+1}=\frac{s_n}{\sqrt{c}}-bs_n^{3/2}c^{(3n-2)/4}$$
$$s_{n+1}=s_n\left(c^{-1/2}-bs_n^{1/2}c^{(3n-2)/4}\right)$$
To contradict $\,s_\infty=0$, it suffices to show:
$$\exists{m}\,\forall{n \geq m}\,\left(s_n\leq1\implies s_{n+1}>s_n\right)$$
This essentially says: For large enough $\,n$, $\,s_n\,$ cannot decrease much further than $\,1$. Hence $\,s_n\,$ cannot converge to $\,0$.
$$s_{n+1}>s_n \iff c^{-1/2}-bs_n^{1/2}c^{(3n-2)/4}>1 \iff c^{1/2}+bs_n^{1/2}c^{3n/4}<1$$
$$s_n\leq1 \implies c^{1/2}+bs_n^{1/2}c^{3n/4} \leq c^{1/2}+bc^{3n/4}$$
So it comes down to choosing an $\,m\,$ such that
$$\forall{n \geq m} \quad c^{1/2}+bc^{3n/4}<1$$
Since $\,0<c<1$, it is clearly possible to choose such an $\,m$.
$$\therefore \quad a_\infty\in\left(0, a_0\right)$$
$$\lim_{n\to\infty}\frac{a_n-a_\infty}{c^{n/2}}=\frac{ba_\infty^{3/2}}{1-\sqrt{c}}\neq0$$
$$a_n=a_\infty+\Theta(c^{n/2})$$
We also have bounds on $\,a_\infty$:
$$\forall{n}\quad\left(a_n-\frac{ba_n^{3/2}c^{n/2}}{1-\sqrt{c}}<a_\infty<a_n\right)$$
We can use the lower bound to prove $\,a_\infty\neq0$. Assume $\,a_\infty=0\,$ for contradiction.
$$a_n-\frac{ba_n^{3/2}c^{n/2}}{1-\sqrt{c}}<0 \iff \sqrt{c}+ba_n^{1/2}c^{n/2}>1$$
Letting $\,n\to\infty$, we get $\,\sqrt{c}\geq1$, which contradicts $\,c<1$.