The series $1 + v^2 + v^4 + v^6 + v^8 + ...$ is gamma squared

Question

Can anyone tell me anything about the series $ 1 + v^2 + v^4 + v^6 + v^8 ... $

Does it have a name? Does it have any special properties?

I ask because $ \gamma = \sqrt{1 + v^2 + v^4 + v^6 + v^8 ...} $

Obviously it only works for v < 1

(i.e. less than the speed of light)

edit: hmmm...subtract one and divide by v^2 and the result is itself

edit: oh and $ \gamma - 1 \approx v^2/2 $ for small values hence

kinetic energy = $m (\gamma - 1) c^2 \approx ( E = m v^2/2$)

Answer 1

Taking $v$ to be the speed of an object relative to some inertial frame $O$, $\gamma = (1-v^2/c^2)^{-1/2}$, then by the geometric series formula:

$$1+\frac{v^2}{c^2} + \frac{v^4}{c^4} + \ldots = \frac{1}{1-\frac{v^2}{c^2}}=\gamma^2$$

Therefore if $v$ in your question is in natural units, you have $1+v^2+v^4 + \ldots = \gamma^2$.

Answer 2

This is a question about special relativity, in which $v$ (often denoted $\beta$) satisfies $\gamma=1/\sqrt{1-v^2}$. As @ziprirovich notes, it's a geometric series, which converges to $\frac{1}{1-v^2}=\gamma^2$ because $|v|<1$.