If there are $n$ frames and the $i$th frame has velocity $v$ with respect to $i-1$th frame. How do I derive the relation between velocity in $S_0$ and $S_n$ frame?
I found velocity in nth frame to be $u_n=\gamma^nu_0-v(\sum_{i=1}^n\gamma)$
What happens when n tends to infinity?
Here $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
Suppose you have your base frame $S_0$ and a frame $S_1$ moving at a relative speed $v_1$. Then you have a second frame $S_2$ moving at $v_2$ relative to $S_1$. To calculate the speed of $S_2$ relative to your base frame you use the equation for the relativistic addition of velocities:
$$ v_{02} = \frac{v_1 + v_2}{1 + \frac{v_1v_2}{c^2}} \tag{1} $$
You could then use this to calculate $v_{02}$, then use it again to calculate $v_{03}$, and so on though that is rapidly going to get tedious. This is where the concept of rapidity mentioned by Ken G comes in.
Firstly let's write all our velocities as a fraction of the speed of light, $v/c$, in which case equation (1) simplifies to:
$$ v_{02} = \frac{v_1 + v_2}{1 + v_1v_2} \tag{2} $$
Now suppose we take the inverse hyperbolic tangent of this. This seems a strange thing to do, but you'll see why this simplifies things. The atanh function is:
$$ \text{atanh}(x) = \tfrac{1}{2}\ln\left(\frac{1+x}{1-x}\right) $$
If we take the atanh of equation (2) we get:
$$ \text{atanh}(v_{02}) = \tfrac{1}{2}\ln\left(\frac{1+\frac{v_1 + v_2}{1 + v_1v_2}}{1-\frac{v_1 + v_2}{1 + v_1v_2}}\right) $$
This apparently horrendous equation simplifies very easily. We just multiply everything inside the $\ln$ by $1 + v_1v_2$ and gather terms and we get:
$$\begin{align} \text{atanh}(v_{02}) &= \tfrac{1}{2}\ln\left(\frac{(1+v_1)(1 + v_2)}{(1 - v_1)(1 - v_2)}\right) \\ &= \tfrac{1}{2}\ln\left(\frac{1+v_1}{1 - v_1}\right) + \tfrac{1}{2}\ln\left(\frac{1 + v_2}{1 - v_2}\right) \\ &= \text{atanh}(v_1) + \text{atanh}(v_2) \end{align}$$
So the $\text{atanh}$ of $v_{02}$ is calculated just by adding the $\text{atanh}$s on $v_1$ and $v_2$. Calculating the relative velocity for the third frame just means adding $\text{atanh}(v_{03})$:
$$ \text{atanh}(v_{03}) = \text{atanh}(v_1) + \text{atanh}(v_2) + \text{atanh}(v_n) $$
And it should be obvious the general case is:
$$ \text{atanh}(v_{0n}) = \sum_{i = 1}^n \text{atanh}(v_i) $$
The quantity $\text{atanh}(v)$ is called the rapidity, and this is what Ken means when he says the rapidities just add together.
The reason why we get this surprising behaviour is that a frame $S_1$ moving at $v_1$ is related to our rest frame by a hyperbolic rotation of a hyperbolic angle $\theta_1 = \text{atanh}(v_1)$. A second frame $S_2$ moving at $v_2$ relative to $S_1$ is rotated relative to $S_1$ by $\theta_2 = \text{atanh}(v_2)$, and the angles of rotation just add. So relative to use it is rotated by:
$$ \theta_{02} = \theta_1 + \theta_2 = \text{atanh}(v_1) + \text{atanh}(v_2) $$
That's why the rapidities add.
It sounds like you are imagining a chain of reference frames, each that perceives the next one as moving with v. So nonrelativistically, the nth frame would move at nv relative to the 0th frame, but of course that fails if nv ~ c. The best way to add a series of frames is to use the concept of rapidity (https://en.wikipedia.org/wiki/Rapidity), so instead of saying each frame has speed v, say each frame has rapidity atanh(v/c) (that's the inverse hyperbolic tangent) relative to the previous one. The advantage is, the rapidities are what simply add, so the rapidity of the nth frame is n times atanh(v/c). So to get the velocity u of the nth frame, say atanh(u/c) = n atanh(v/c), or u = c*tanh(n atanh(v/c)).
