Is there a closed form for the recurrence $V_{n+1}={V_n+\Delta V\over 1+{V_n\cdot \Delta V/C^2}}$, for constants $\Delta V$ and $C$?
Defining $U_n := V_n/C$ and $D := \Delta V/C$, we can write the simpler-looking recurrence $$U_{n+1} = \frac{D + U_n}{1 + D\,U_n} \tag{1}$$ which bears a resemblance to the angle-addition formula for hyperbolic tangent: $$\tanh(a+b) = \frac{\tanh a + \tanh b}{1 + \tanh a\,\tanh b} \tag{2}$$ Thus, if we further define $u_n := \operatorname{arctanh U_n}$ and $d := \operatorname{arctanh D}$, then we have $$\tanh u_{n+1} = U_{n+1} = \frac{D+U_n}{1+D\,U_n} = \frac{\tanh d + \tanh u_n}{1 + \tanh d\tanh u_n} = \tanh(d+u_n) \tag{3}$$ so that $u_{n+1} = d + u_n$. Consequently, taking $V_0 = U_0 = u_0 = 0$, we have $u_n = n d$, which gives
$$V_n= C\,\tanh\left( n \operatorname{arctanh}\frac{\Delta V}{C} \right) \tag{$\star$}$$
As $n$ grows without bound (with $\Delta V/C > 0$), the $\tanh$ factor approaches $1$, so that $V_n$ approaches $C$ (as OP has observed).
Using the exponential definition of $\tanh$, we can re-write $(\star)$ as $$V_n = C \frac{e^{nd}-e^{-nd}}{e^{nd}+e^{-nd}} = C\frac{\left(e^d\right)^n-\left(e^d\right)^{-n}}{\left(e^d\right)^n+\left(e^d\right)^{-n}} \tag{4}$$ Note that $$e^d = \exp \operatorname{arctanh} D = \exp \left(\frac12\,\log\frac{1+D}{1-D}\right) = \sqrt{\frac{1+D}{1-D}} \tag{5}$$ Substituting $(5)$ into $(4)$ and simplifying ultimately gives
$$V_n = C\frac{(1+D)^n-(1-D)^n}{(1+D)^n+(1-D)^n} = C\frac{(C+\Delta V)^n-(C-\Delta V)^n}{(C+\Delta V)^n + (C-\Delta V)^n} \tag{$\star\star$}$$
which agrees with @Hushus46's answer.
As I have noted in the comment, Wolfram Alpha gives the closed form, with $k = \Delta V$:
$$V_n = \frac{C_1 c(c+k) \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - c(c-k)\left(-\frac{k}{c(c+k)}\right)^n}{(c-k)\left(-\frac{k}{c(c+k)}\right)^n + C_1 (c+k) \left(\frac{1}{k-c}+\frac{1}{c}\right)^n}$$
Taking $V_0 = 0$, we can see
$$ 0 = \frac{C_1 c(c+k) - c(c-k)}{(c-k) + C_1 (c+k) }$$
which is undefined if $C_1 = -\frac{c-k}{c+k}$, so let $C_1 \ne -\frac{c-k}{c+k}$, then
$$0 = C_1 c(c+k)-c(c-k)\Rightarrow 0 = C_1(c+k)-c+k \Rightarrow C_1=\frac{c-k}{c+k}$$
Hence $C_1$ is undefined when $k = \Delta V = -C$, so depending on the physical context of special realtivity (which I don't know enough of), this may never happen
Substituting this into the original closed form: \begin{align}V_n &= \frac{(\frac{c-k}{c+k}) c(c+k) \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - c(c-k)\left(-\frac{k}{c(c+k)}\right)^n}{(c-k)\left(-\frac{k}{c(c+k)}\right)^n + (\frac{c-k}{c+k}) (c+k) \left(\frac{1}{k-c}+\frac{1}{c}\right)^n} \\ &=\frac{c(c-k)\left[ \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - \left(-\frac{k}{c(c+k)}\right)^n\right]}{(c-k)\left[\left(-\frac{k}{c(c+k)}\right)^n + \left(\frac{1}{k-c}+\frac{1}{c}\right)^n\right]}\\ &= c \frac{\left[ \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - \left(-\frac{k}{c(c+k)}\right)^n\right]}{\left[\left(-\frac{k}{c(c+k)}\right)^n + \left(\frac{1}{k-c}+\frac{1}{c}\right)^n\right]}\end{align}
Hence in terms of the way you formulated you have
$$ \boxed{V_n=C \frac{\left[ \left( \frac{1}{\Delta V-C} + \frac{1}{C}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(\frac{1}{\Delta V-C}+\frac{1}{C}\right)^n\right]} \quad \Delta V \ne \pm C ,\, C \ne 0}$$
if one wishes a preferable altenartive form without negatives inside the power terms:
\begin{align}V_n&=C \frac{\left[ \left( \frac{\Delta V}{C(\Delta V - C)}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(\frac{\Delta V}{C(\Delta V - C)}\right)^n\right]} \\ &=C \frac{\left[ \left( -\frac{\Delta V}{C(C-\Delta V)}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(-\frac{\Delta V}{C(C- \Delta V)}\right)^n\right]} \\ &=C \frac{\left[ (-1)^n \left(\frac{\Delta V}{C}\right)^n\left( \frac{1}{(C-\Delta V)}\right)^n - (-1)^n \left(\frac{\Delta V}{C}\right)^n \left(\frac{1}{(C+\Delta V)}\right)^n\right]}{\left[(-1)^n \left(\frac{\Delta V}{C}\right)^n\left(\frac{1}{(C+\Delta V)}\right)^n + (-1)^n \left(\frac{\Delta V}{C}\right)^n \left(\frac{1}{(C- \Delta V)}\right)^n\right]} \end{align} which is written \begin{align}V_n&=C \frac{\left[ \left( \frac{1}{C-\Delta V}\right)^n - \left(\frac{1}{C+\Delta V}\right)^n\right]}{\left[\left( \frac{1}{C-\Delta V}\right)^n + \left(\frac{1}{C+\Delta V}\right)^n\right]} \\ &=C \frac{\left[ \left( \frac{1}{C-\Delta V}\right)^n - \left(\frac{1}{C+\Delta V}\right)^n\right]}{\left[\left( \frac{1}{C-\Delta V}\right)^n + \left(\frac{1}{C+\Delta V}\right)^n\right]} \frac{(C+\Delta V)^n (C-\Delta V)^n}{(C+\Delta V)^n (C-\Delta V)^n}\end{align} Finally, $$\boxed{V_n=C \frac{\left[ \left( C+\Delta V\right)^n - \left(C- \Delta V\right)^n\right]}{\left[\left( C+ \Delta V\right)^n + \left(C- \Delta V\right)^n\right]} \quad \Delta V \ne \pm C ,\, C \ne 0}$$
which is in the form:
$$V_n = C \frac{a^n-b^n}{a^n+b^n}$$
which has limit $C$ as $n \to \infty $ (see this if you don't know how to show that), matching the limit in the OP's question.
If anyone knows how to derive the solution Wolfram gets, that would be preferable to this