So I'm having trouble at figuring out how to calculate the MGF of the geometric function for the probability density function of $f(x)=\frac1{2^x}$ where x is discrete.
I know that $$M(t)=\sum_{x=1} e^{tx} \frac{1}{2^x}, x=1,2,3,..$$
and that the desirable answer is
$$M(t) = \frac{e^t}{2-e^t}$$, but the series looks to me as if it was growing one and I'm unable to figure out the correct way to start off.
Rewrite $$ \sum_{x=1}^\infty e^{tx} \frac{1}{2^x} = \sum_{x=1}^\infty \left(\frac{e^t}{2}\right)^x $$ which is a geometric series with ratio $e^t/2$. Using the standard formula "$\frac{first\ term}{1 - ratio}$" you get $$ \sum_{x=1}^\infty \left(\frac{e^t}{2}\right)^x = \frac{\frac{e^t}{2}}{1-\frac{e^t}{2}} $$ Then multiplying above and below by $2$ gives you the result.
(This is valid for $t < \ln 2)$