Relativity

For motion in one dimension, if $U$ is the speed of $B$ relative to $A$ , while $V$ is the speed of $C$ relative to $B$ , then the speed of $C$ relative to $A$ is $W$ , where $\tanh^{-1}\tfrac{W}{c} \; =\; \tanh^{-1}\tfrac{U}{c} + \tanh^{-1}\tfrac{V}{c}$ Thus, for this problem. we have $\tanh^{-1}\tfrac{v}{c} \; = \; \displaystyle \sum_{n=1}^\infty \tanh^{-1}2^{-n} \; = \; \tfrac12\sum_{n=1}^\infty \ln\left(\frac{1 + 2^{-n}}{1 - 2^{-n}}\right) \; = \; \displaystyle \tfrac12\ln\left(\prod_{n=1}^\infty \frac{1+2^{-n}}{1 - 2^{-n}}\right) \; = \; \tfrac12\ln\left(\frac{(-1,\frac12)_\infty}{2(\frac12,\frac12)_\infty}\right)$ using the $q$ -Pochhammer symbols. Thus $\tfrac{v}{c} \; = \; \tanh\left(\tfrac12\ln\left(\frac{(-1,\frac12)_\infty}{2(\frac12,\frac12)_\infty}\right)\right) \; =\; \frac{(-1,\frac12)_\infty - 2(\frac12,\frac12)_\infty}{(-1,\frac12)_\infty + 2(\frac12,\frac12)_\infty} \; = \; \boxed{0.783924...}$

The formula for relativistic velocity addition is: $v^\prime=\frac{u+v}{1+\frac{uv}{c^2}}$

To make things less messy, let $a_n=\frac{v_n}{c}$ . Thus, we can derive that

$a_1=\frac{1}{2}\\a_n=\frac{a_{n-1}+\frac{1}{2}^n}{1+(a_n-1)\frac{1}{2}^n}$

This rapidly converges to 0.7839236595945458.