Specific questions are bolded below. I've been unsuccessful in solving the following problem., which is exercise 5.2.3 from Probability and Random Processes by Grimmett and Stirzaker.
Let $X_1, X_2,\ldots$ be independent and identically distributed random variables with the logarithmic mass function $ f(k) = \frac{(1-p)^k}{k \log(1/p)}, \quad k \geq 1, $
where $0 < p < 1$. If $N$ is independent of the $X_i$ and has the Poisson distribution with parameter $\mu$, show that $Y=\sum_{i=1}^N X_i$ has a negative binomial distribution.
My strategy is to compute the probability generating function (pgf) for $Y$ and and the pgf for a negative binomial distribution (nbd) and show that they're the same, which would imply that $Y$ is indeed distributed negative binomially. I'm using the fact that random variables $f$ and $g$ have the same distribution iff they have the same pgf. Is it true that random variables $f$ and $g$ have the same distribution iff they have the same pgf? If so, please continue reading my attempted solution.
Since $N$ is independent of each $X_i$, Theorem 5.1.25 of the textbook says that $G_Y= G_{N} \circ G_{X}.$
It's easy to compute (and it's done in the textbook) $G_{N}(s) = \exp(\mu(s-1)).$ Further $ \begin{align} G_{X}(s) &= \sum_{k\geq 1}s^{k} \frac{(1-p)^k}{k \log(1/p)} \\ &= \frac{1}{\log(1/p)} \sum_{k\geq 1} \frac{(s(1-p))^k}{k} \\ &= \frac{\log(1/q)}{\log(1/p)} \sum_{k\geq 1} \frac{(1-q)^k}{k\log(1/q)} \qquad (q := 1-s(1-p))\\ &= \frac{\log(1/q)}{\log(1/p)}\qquad \bigg(\text{Since } \frac{(1-q)^k}{k\log(1/q)} \text{ is a pmf.}\bigg) \\ &= \frac{\log q}{\log p} \\ &= \frac{\log(1-s(1-p))}{\log p}. \end{align} $ Therefore $ \begin{align} G_{Y}(s) &= G_{N}(G_{X}(s)) \\ &= \exp\bigg(\mu \big(\frac{\log(1-s(1-p))}{\log p} - 1\big)\bigg). \end{align} $
Now since an nbd with parameters $r$ and $\alpha$ is a sum of $r$ independent geometric random variables (each with parameter $\alpha$), that nbd has pgf given by raising the pgf of each geomtric random variable to the power $r$: $ \left(\frac{\alpha s}{1-s(1-\alpha)}\right)^{r}. $ However, setting $ \exp\bigg(\mu \big(\frac{\log(1-s(1-p))}{\log p} - 1\big)\bigg) = \left(\frac{\alpha s}{1-s(1-\alpha)}\right)^{r} $ allows me to solve for $\alpha$, but gives something weird, which I doubt is correct. In fact, I suspect that it should be the case that $Y$ is nbd with paramters $\alpha = p$ and $r= \mu$.
Where am I erring?
Is there a better (i.e. simpler) approach to this problem?
As a side question, I noticed that Wikipedia gives the pgf of a negative binomial as $ \left(\frac{1-\alpha}{1-\alpha s}\right)^{r}, $ which is apparently different from the one I calculated above (though still not clearly equal to $G_{Y}(s)$ for any values of $\alpha$ and $r$). Why the discrepancy between my calculation of the pgf of an nbd and Wikpedia's entry?
EDIT In the textbook I cited above, a random variable $W_{r}$ is defined to be nbd if it has pmf $ \mathbb{P}(W_{r} = k) = \binom{k-1}{r-1} \alpha^{r} (1-\alpha)^{k-r}, \qquad k=r,r+1,\ldots. $ It is then pointed out that $W_{r}$ is the sum of $r$ independent geometric random variables; i.e. $ W_{r} = Z_{1} + Z_{2} + \cdots + Z_{r}, $ where each $Z_{i}$ has pmf $ f_{Z} = \alpha(1-\alpha)^{k-1}, \qquad k=1,2,\ldots. $