Proof for moment generating function for $t<-\ln(1-p)$ 
Question: Let X be a random variable with the geometric distribution $f(x) = p(1-p)^{x-1}$
Obtain the moment generating function for $t<-ln(1-p)$

Hint: $\sum_{k=0}^{\infty}r^{k} = \frac{1}{1-r}, |r|<1$
Proof:
Let $M_{X}(t)$ be the moment generating function for a random variable $X$.
Then, $M_{X}(t) = E[e^{Xt}] = \sum_{x=1}^{\infty}e^{xt}p(1-p)^{x-1} = p\sum_{x=1}^{\infty}e^{xt}(1-p)^{x-1}$
I would appreciate an obvious hint to proceed me through.
Thanks in advance.
 A: Now you have to apply the formula for the partial sum of a geometric series (given hint): 
$$\sum_{k=1}^n r^k=r\cdot \frac{r^n-1}{r-1}$$
For $|r|<1$ the series converges: $\sum\limits_{k=1}^{\infty} r^k=\lim\limits_{n \to \infty }r\cdot \frac{r^n-1}{r-1}=r\cdot \frac{0-1}{r-1}=\frac{r}{1-r} \qquad (*)$
Next we we simplify the sum:
$$M_{X}(t)  = p\sum_{x=1}^{\infty}e^{xt}(1-p)^{x-1}=p\sum_{x=1}^{\infty}\left(e^{t}\right)^x(1-p)^{x-1}$$
$$=p\sum_{x=1}^{\infty}\left(e^{t}\right)^x(1-p)^{x}\cdot \frac1{1-p}=\frac{p}{1-p}\sum_{x=1}^{\infty}\left(e^{t}\right)^x(1-p)^{x}=\frac{p}{1-p}\sum_{x=1}^{\infty}(e^{t}(1-p))^{x}$$
The series converges if $e^{t}(1-p)<1$. We can check that. The given condition is that $t<-\ln(1-p)$. Multiply the inequality by $(-1)$:$-t>\ln(1-p)$. The inequality sign turns around.  Writing both sides as exponents of e: $e^{-t}>1-p$
Multiplying both sides by $e^{t}$
$$e^{-t}\cdot e^t>(1-p)\cdot e^t\Rightarrow 1>(1-p)\cdot e^t \qquad \color{blue}{\checkmark}$$
To obtain $M_X(t)$ you finally replace $r$ by $e^{t}(1-p)$ at $\Large{\frac{p}{1-p}\cdot \frac{r}{1-r}}$ 
I think you can finish...
A: Note that
$$
\begin{align}
p\sum_{x=1}^{\infty}e^{xt}(1-p)^{x-1}
&=pe^t\sum_{x=1}^{\infty}e^{(x-1)t}(1-p)^{x-1}\\\tag{0}
&=pe^t\sum_{u=0}^{\infty}e^{ut}(1-p)^{u}\\
&=pe^t\sum_{u=0}^{\infty}(e^t(1-p))^{u}\\
&=\frac{pe^t}{1-e^t(1-p)}\tag{1}
\end{align}
$$
provided that $|e^t(1-p)|=e^t(1-p)<1$ where in (0) we made the change of variables $u=x-1$ and in $(1)$ we use the hint with $r=e^t(1-p)$.
