All details are in the attached picture. Please refer to it.
Here are some examples I calculated with different first year operating margins(in case it might help you):
Example #1: When terminal year operating margin = 14.18% and my speed of convergence = 1:
Yr 0 = -4.18%, Yr 1 = 5%, Yr 2 = 9.6%, Yr 3 = 11.9%, Yr 4 = 13.04%, Yr 5 = 13.62%, Yr 6 = 13.91%, Yr 7 = 14.05%, Yr 8 = 14.12%, Yr 9 = 14.16%, Yr 10 = 14.18%, Yr 11 (Terminal value) = 14.19%
Example #2: When terminal year operating margin = 100% and my speed of convergence = 1:
Yr 0 = 100%, Yr 1 = 57.10%, Yr 2 = 35.65%, Yr 3 = 24.92%, Yr 4 = 19.56%, Yr 5 = 16.88%, Yr 6 = 15.53%, Yr 7 = 14.86%, Yr 8 = 14.53%, Yr 9 = 14.36%, Yr 10 = 14.28%, Yr 11 (Terminal value) = 14.19%
Also, I want to know how can one make better formulas that convergent to a given value. Does anyone have any appropriate resources/methodology they could share?
You are doing 11 times the average between a number and $14.19$: calling $x_n$ the operating margin at Year $n$, you have
$$x_n=\frac{1}{2}x_{n-1}+\frac{14.19}{2}=\frac{1}{4}(x_{n-2}+14.19)+\frac{14.19}{2}=\frac{1}{4}x_{n-2}+\frac{3}{4}14.19=\\ =\dots=\frac{1}{2^n}x_0+(1-\frac{1}{2^n})14.19.$$
Considering the fact that $2^{11}=2048$, and you consider only 2 decimals you have $$\left(1-\frac{1}{2^{11}}\right)14.19=14.18.$$
Moreover, choosing $x_0\in(10,100)$, $\frac{1}{2^{11}}x_0\in(0.0005,0.005)$ which is almost enough to reach $14.19$, and you have also rounding errors that sums up, so it is quite understandable that the convergence seems perfect.
If you try with huge numbers, it won't be the case. Indeed, for $x_0=2048$, you will have $$x_{11}=1+\frac{2047}{2048}14.19=1+14.18=15.18.$$
In particular, choose $a$ to be any number, and take $x_0=2048\,a+14.19$, hence $x_{11}=a$.
EDIT
The general case, can be written as follows: let $x_0$ be the initial operating margin, $s$ the speed of convergence, hence
$$x_n=\frac{s}{1+s}x_{n-1}+\frac{14.19}{1+s}=\frac{s}{(1+s)^2}(s\,x_{n-2}+14.19)+\frac{14.19}{1+s}=\\ \left(\frac{s}{1+s}\right)^2 x_{n-2}+\left(1-\frac{s^2}{(1+s)^2}\right)14.19=\dots=\\ =\left(\frac{s}{1+s}\right)^n x_0+\left(1-\left(\frac{s}{1+s}\right)^n\right)14.19.$$
Since $\left(\frac{s}{1+s}\right)^n\to0$, it is clear that for $n$ big enough, it converges to $14.19$, the fact is that depending on the choice of $x_0$ and $s$, big enough can be bigger or smaller than 11, which is the number you want.
In particular, with the equation above, putting n=11, you can play with $s$ and $x_0$ and see how far you are from the convergence.