In my question I have $M_X(t)= 5/(3-e^t)$, suppose I want to calculate $P(X<3)$.
From the book that I am using, I usually can read off the values of the $M_X(t)$ i.e. if $M_X(t)=1e^t$ then $P(X=1)=1$, however here my $e^t$ is in the denominator position, can someone point me in the right direction here? I have tried rationalizing the fraction but it was not of any use.
Hint: Use geometric expansion. Notice that $$\frac{5}{3-e^t}=\frac{5}{3}\cdot \frac{1}{1-\frac{e^t}{3}}=\frac{5}{3}\sum _{k=0}^{\infty}\frac{1}{3^k}e^{k\cdot t}.$$ Can you finish?