Calculate the Laplace transform Help me calculate the Laplace transform of a geometric series.
$$
 f(t) = \sum_{n=0}^\infty(-1)^nu(t-n)
$$
show that 
$$
\mathcal{L} \{f(t)\} = \frac{1}{s(1+\mathcal{e}^{-s})}
$$
Edit:
 so far I know that 
$$
\mathcal{L} \{f(t)\} = \frac{1}{s}\sum_{n=0}^\infty(-1)^ne^{-ns}
$$
 A: Here is a simpler approach.
Note that for $t>0$, we have $f(t)+f(t-1) = 1$. The Laplace transform of a time-shifted function (in this case $f(t-1) = f(t-1) u(t-1)$) is $s \mapsto e^{-s} \hat{f}$, where $\hat{f}$ is the Laplace transform of $f$. Furthermore, the Laplace transform of $1$ is just $s \mapsto \frac{1}{s}$. Hence we have
$$\hat{f}(s)+ e^{-s} \hat{f}(s) = \frac{1}{s},$$
from which it follows that
$$\hat{f}(s) = \frac{1}{s(1+e^{-s})}.$$
(Your formula above is incorrect.)
A: Try to think what your sum looks like. You will see it is a periodic function, with period $2$, that we can define as.
$$f(t) = 1 ; 0<t<1$$
$$f(t) = 0 ; 1<t<2$$
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
$$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\int_0^2 e^{-st}f(t)dt$$
$$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\int_0^1e^{-st}dt$$
$$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\frac{1-e^{-s}}{s}$$
$$\mathcal L \{ f(t) \} =\frac{1}{1+e^{-s}}\frac{1}{s}$$
A: A two-step solution:
Fact 1. $L\big[H_n(t)]=\frac{e^{-pn}}{p};$
Fact 2. $\sum_0^{\infty}(-1)^n\frac{e^{-pn}}{p}=\frac{1}{p}\frac{1}{1+e^{-p}},\;\;
Rep>0.$
Here, $H_n(t)=u(t-n).$
