Try to think what your sum looks like. You will see it is a periodic function, with period $2$, that we can define as.
$f(t) = 1 ; 0
$f(t) = 0 ; 1
In general if a function has period $P$, it's LP is given by
$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-sP}}\int_0^P e^{-st}f(t)dt$
under appropriate conditions (i.e. the LP should exist)
It this case $f(t)$ is periodically constant, so the LP exists, and
$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\int_0^2 e^{-st}f(t)dt$
$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\int_0^1e^{-st}dt$
$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\frac{1-e^{-s}}{s}$
$\mathcal L \{ f(t) \} =\frac{1}{1+e^{-s}}\frac{1}{s}$