I am trying to solve the following:
The nonnegative, integer-valued, random variable X has generating function $g_{X(t)}=log(\frac{1}{1-qt})$. Determine $P(X = k)$ for $k = 0, 1, 2, . . .$, $E[X]$, and $Var[X]$.
I know the probability generating function is of the form $g_{X(t)}=E[t^{x}]=\sum_{n=1}^{\infty}t^{n}*P(X=n)$. I thought I could do a Taylor Series expansion of the $g_{X(t)}$ and make maniuplate it into the form of $\sum_{n=1}^{\infty}t^{n}*P(X=n)$, but I'm not sure I can get it to factor. When I take the Taylor Series expansion I get $log(\frac{1}{1-qt}) = -\sum_{k=1}^{\infty} \frac{(-1)^{k} (\frac{qt}{1-qt})^k}{k}$ for $|{\frac{x}{1-x}}|<1$. I'm not sure what to do about the alternating term, and this doesn't look like anyone of the discrete distributions I am familiar with.
To get the variance and expected value I should be able to use $E[X]=g'_x(1)$ and $Var[X]=g''_x(1)+g'_x(1)-(g'(1))^2$, where I substitute $qt=1$ into the probability generating function.
You've got the wrong Taylor series.
$$\log\left(\frac{1}{1-z}\right)=\sum_{n=1}^\infty \frac{z^n}{n}$$
for $|z|<1$.
Setting $z=qt$ gives you $P(X=n)=\frac{q^n}{n}$.
Note that this is not a probability function for all $q$, because you'd need $g_X(1)=1$, that is $\log\left(\frac{1}{1-q}\right) = 1$, which means you can only have one specific $q=1-e^{-1}$, or you need to scale $\log\left(\frac{1}{1-qt}\right)$.