What is the moment generating function of $P(X=x)=\alpha\theta^x$? Let $X$ be a discrete random variable such that $P(X=x)=\alpha\theta^x$, $x=1,2,\ldots,$ and $0 < \theta < 1$ and $P(X=0)=1-\sum^\infty_{x=1}\alpha\theta^x$, where $\alpha$ is a constant. Find the $M_X(t)$ with its appropriate ranges for $t$ and $\alpha$.
My work so far:
$M_X(t)=E(e^{tx})=\sum^\infty_{x=1}\alpha e^{tx}\theta^x= \alpha \sum^\infty_{x=1} e^{tx} \theta^x = \alpha b\theta(1+b\theta^2+b\theta^3+\cdots)$, where $b=e^t$. I just do not know how to manipulate this series so that we get a nice function for $M_X(t)$. Also, I am wondering how you can make use of $P(X=0)$ in the solution.
Thank you.
 A: See @Glen-b's comment. 
What follows finds the MGF for an ordinary geometric distribution (where $\alpha = 1-\theta)$ that
counts the number of failures before the first success, and so
has support on the nonnegative integers. [There are other versions
of the geometric distribution.]
To make typing easier, I will let $p$ be the success probability and
$q = 1-p.$ Thus $P(X = k) = pq^k,$ for $k = 0, 1, \dots.$ Then
the MGF $M_X(t)$ is found as follows:
$$M_X(t) = \sum_{k=0}^\infty e^{tk}pq^k = 
p\sum_0 (qe^t)^k = \frac{p}{1-qe^t},$$
where we have summed an infinite series at the last equality,
subject to the restriction that $qe^t < 1.$ [You should express
this in terms of $t$ when stating the values of $t$ for which
the MGF exists.]
Knowing the form of $P(X=0)$ makes it easy to include the $0$th
term into the series from the start.
Ref: You may want to look at Wikipedia where both of
the most commonly used versions of the geometric distribution are
shown (we're using their second version). Also useful may be 
this page,
for its clear exposition, even though it uses the alternate version of
the PDF. Finally, this page from a sister site finds the mean of a geometric distribution without using the MGF.
Addendum: You roll a fair die until the first 6 occurs. The number $X$ of failures before that success, it our geometric random variable. Find $E(X).$ First, $M_X^\prime(t) = pqe^t/(1-qe^t)^2.$
Then $M_X^\prime(0) = q/p = \frac{5/6}{1/6} = 5.$
This is the version of the geometric distribution implemented in R.
x = 0:500;  p = 1/6;  q = 5/6  
pdf=p*q^x;  sum(x*pdf)
[1] 5             # summing 500 terms suffices for E(X)
sum(x*dgeom(x,p))
[1] 5             # same thing, using R's function 'dgeom'
set.seed(916);  mean(rgeom(10^6, 1/6))
[1] 4.99737       # avg of 10^6 simulated 'games' for 2-place accuracy

