Showing that $\mathbb{P}(X_1=k) = \frac{(1-p)^k}{k\log(1/p)}\quad k\in\mathbb{N}$ is a distribution on $\mathbb{N}$

Question

Let $X_1, X_2,\ldots$ be independent and identically distributed random variables with the logarithmic mass function $$ \mathbb{P}(X_1=k) = \frac{(1-p)^k}{k \log(1/p)}, \quad k \in \mathbb{N} \quad (1) $$

,where $0 < p < 1$.

(i) Show that (1) is a distribution on the Set $\mathbb{N}$.

(ii) Let $N$~Poi$(\alpha)$ with the parameter $\alpha>0$. Show that $$ S_N := \sum_{i=1}^{N}X_i $$ is negatively binomially distributed if $N$ and $(X_n)_{n\in\mathbb{N}}$ are independent.

I am somewhat confused by what I am meant to do for (i) my idea is to compute the Probability Generating Function(pdf), but after doing so I am not sure how to argue why it in fact is a distribution. For question (ii) I am quite sure I could solve it using Probability Generating Functions, as well as their composition.

Answer 1

To prove that $(1)$ is a distribution you need to show $$\sum_{k=1}^\infty P(X_1=k) = 1$$

or equivalently that

$$\sum_{k=1}^\infty \frac{(1-p)^k}{k} = \log(1/p)$$

Now write $\log(1/p)=-\log (p)=-\log(1+(p-1))$. Since $|p-1|\leq 1$ we can use the Taylor series expansion for $\log(1+x)$ to get

$$\log(1/p) = -\sum_{k=1}^\infty\frac{(-1)^{k+1}(p-1)^k}{k} = \sum_{k=1}^\infty \frac{(p-1)^k}{k}$$ thus proving that $(1)$ is actually a distribution on $\mathbb{N}$.