how to normalize discrete probability distribution $P_n = \binom{N + n - 1}{n} p^N (1 - p)^n$ How do I normalize this distribution?
$P_n = \binom{N + n - 1}{n} p^N (1-p)^n$
I know that $\sum_{n=0}^{\infty} (1-p)^n = \frac{1}{1-(1-p)}$.

Also I know that $\sum_{N=0}^{\infty} \binom{N + n - 1}{n - 1} p^N = \frac{1}{(1-p)^n}$

And I know that eventually the cumulative result of $P_n$ must be $1$.
But I don't know how to handle the $\binom{N + n - 1}{n} p^N$.
Furthermore, how would I calculate the expected value of this distribution?
 A: According to this, the distribution is already normalized:
$$\sum_{n \in \mathbb{N}_0}\binom{N+n-1}{n}p^N(1-p)^n=p^N\frac{1}{(1-(1-p))^N}=1$$
I also checked it with a script:
N = 20
M = int(1e+04) 
p = 0.67

Pn = 0
for n in range(0,M):
    Pn += scipy.special.comb(N+n-1,n)*p**N*(1-p)**n
    
print(Pn)


Also the expected value is
$$\mathbb{E}[X]=N\frac{1-p}{p}$$
Indeed, we can compute
$$\begin{align*}
\mathbb{E}[X] &= \sum_{n=0}^\infty n P_n
= \sum_{n=1}^\infty n \binom{N+n-1}{n}p^N(1-p)^n \\
&= \sum_{n=0}^\infty (n+1) \binom{N+n}{n+1}p^N(1-p)^{n+1} \\
&= \frac{1-p}{p}\sum_{n=0}^\infty (n+1) \binom{N+n}{n+1}p^{N+1}(1-p)^{n} \\
&= \frac{1-p}{p}\sum_{n=0}^\infty (n+1) \frac{(N+n)!}{(n+1)!(N-1)!}p^{N+1}(1-p)^{n} \\
&= N\frac{1-p}{p}\sum_{n=0}^\infty \frac{(N+n)!}{n!N!}p^{N+1}(1-p)^{n} \\
&= N\frac{1-p}{p}\sum_{n=0}^\infty \binom{(N+1)+n-1}{n}p^{N+1}(1-p)^{n} \\
&= N\frac{1-p}{p}\cdot 1
\end{align*}$$
since the last series is $1$, by the above argument (it is the original sum of probabilities, but with $N+1$ as parameter instead of $N$).
