How to find the MGF of this pmf?

Question

A random variable $X$ has pmf $p(x;\alpha) = (1-\alpha)^{x-1} \alpha$ for $x = 1,2,\dots$. Find the moment generating function of $X$, $M_X(t)$.

What I've done:

$E[e^{tx}] = \sum_x e^{tx} (1-\alpha)^{x-1} \alpha = \frac{\alpha}{1-\alpha} \sum_x (e^t (1-\alpha))^x$.

Assume $e^t(1-\alpha) < 1$. Then the series is geometric with $a_0 = 1$, $r < 1$ hence the sum is

$\frac{a_0}{1-r} = \frac{1}{(1-\alpha)}\frac{1}{1-(e^t(1-\alpha))}$

But this is wrong. Answer should be $\frac{\alpha e^t}{1-(1-\alpha)e^t}$. Where have i erred?

Answer 1

Since $x$ starts at $1$, your initial term should be $e^t(1-\alpha)$. Then we get the MGF as $$\frac{\alpha}{1-\alpha}\frac{e^t(1-\alpha)}{1-e^t(1-\alpha)}=\frac{\alpha e^t}{1-(1-\alpha)e^t},$$as expected.

Answer 2

$$\sum_{x=1}^{\infty}e^{tx}(1-\alpha)^{x-1}\alpha=\alpha e^t\sum_{x=0}^{\infty}e^{tx}(1-\alpha)^x=\frac{\alpha e^t}{1-e^t(1-\alpha)}$$

Your mistake is overlooking that $x$ starts at $1$ (hence not at $0$).