Why is this a geometric distribution?

Question

For a random variable $X$, $$P(X = x) = (p-1)/p^{(x + 1)}$$ where $p$ is in $(1,\infty)$. Why is $X$ geometrically distributed? (and why would this make it true that $E[X] = 1 / (p - 1)$ ?) I know a geometric distribution to be $$P(X = x) = p ((1-p)^x)$$ and I do not know how to turn what I have into this. Is $X$ as I defined it above geometrically distributed and if yes why?

Answer 1

For every $p\gt1$, $r=1-1/p$ is in $(0,1)$ and, for every $x$, $(p-1)/p^{x+1}=r(1-r)^{x}$. Thus, the distribution of $X$ is geometric with parameter $r$.

If $p=1$, the identities $P(X = x) = (p-1)/p^{(x + 1)}$ do not define a legitimate distribution since one would have $P(X=x)=0$ for every $x$.