Answer
$\displaystyle \color{blue}{\frac{\frac{1}{4}}{1 - \frac{3}{4}e^t}}$
Work Step by Step
From Example 3.12.1 (p. 206), we see that the given pdf of $X$ is geometric with a probability of success per trial $p=\frac{1}{4}.$ Thus, the mgf of $X$ is immediately obtained to be
$\displaystyle \color{blue}{M_X(t) = \frac{\frac{1}{4}}{1 - \frac{3}{4}e^t}}$
based on the result of this example (see p. 207).
Alternatively, we can use the definition of the mgf and the given pdf to obtain:
$\begin{align*}
M_X(t) &= E(e^{Xt}) \\
&= \displaystyle \sum_{k=0}^\infty e^{kt}p_X(k) \\
&= \sum_{k=0}^\infty e^{kt}\left(\frac{3}{4}\right)^k\left(\frac{1}{4}\right) \\
&= \sum_{k=0}^\infty (e^t)^k\left(\frac{3}{4}\right)^k\left(\frac{1}{4}\right) \\
&= \sum_{k=0}^\infty \left(\frac{3e^t}{4}\right)^k\left(\frac{1}{4}\right) \\
&= \left(\frac{1}{4}\right) \underbrace{\sum_{k=0}^\infty \left(\frac{3e^t}{4}\right)^k}_{S_\infty\ \text{of G.P.},\ t_1\;=\;1,\ r\;=\;\frac{3e^t}{4}} \\
&= \left(\frac{1}{4}\right) \overbrace{\frac{1}{1-\frac{3e^t}{4}}}^{S_\infty = \frac{t_1}{1-r}} \\
\color{blue}{M_X(t)}\ &\color{blue}{= \frac{\frac{1}{4}}{1 - \frac{3}{4}e^t}}
\end{align*}$
