I was given the following exercise.

Let $X$ be a random variable that takes non-negative natural number

values such that $E(X)=1,E(X^2)=2,E(X^3)=5$. Find the minimum value of

$P(X=0)$using the taylor expansion of the probability generating function

at $z=1$.

I know the method not using the generating function stated in this question.

**My attempt:**

Let $G(z)=E(z^X)$ be the probability generating function of $X$. Then by definition,

$G(z) = P(X=0)+P(X=1)z+P(X=2)z^2+cdots $

Also, since $E(frac{X(X-1)cdots (X-n+1)}{n!})=frac{G^{(n)}(1)}{n!}$ according to Wikipedia, we have

$displaystyle G(z) = sum_{n=0}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(z-1)^n$

Substituting $z=0$ yields

$displaystyle P(X=0) = sum_{n=0}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(-1)^n$

Note that for $n=0,1,2,3$, the coefficient can be calculated as follows:

$begin{align} E(1) &= 1 \ E(X)&= 1 \ E(X(X-1)/2) &= E(X^2)/2 -E(X)/2 = 1-1/2 =1/2 \ E(X(X-1)(X-2)/3!) &= E(X^3)/6 -E(X^2)/2 +E(X)/3 = 5/6-2/2+1/3 = 1/6 end{align}$

Therefore,

$displaystyle begin{align} P(X=0) &= sum_{n=0}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(-1)^n \ &=1-1+frac{1}{2} -frac{1}{6} +sum_{n=3}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(-1)^n \ &= frac{1}{3} + sum_{n=3}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(-1)^nend{align}$

According to the question linked above, $1/3$ is the minimum, so I think we need to prove that the sum is non-negative to complete the proof. However, I was unable to do so.

Am I on the right path? If not, what is the correct one? If yes, how can I finish it?