Valid Probability Model problem

Question

I know that a valid probability model must sum to 1 but I am confused on how to check that with this particular sum.

$$P(k)= \frac{(\ln 2)^k}{k!},\qquad k=1,2,3,\ldots $$

Answer 1

Recall the Maclaurin series expansion for the exponential function: $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots=\sum_{k=0}^\infty \frac{x^k}{k!}.$$ Thus $$e^x-1=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots=\sum_{k=1}^\infty\frac{x^k}{k!}.$$ Put $x=\ln 2$. We obtain $$\sum_{k=1}^\infty \frac{(\ln 2)^k}{k!}=e^{\ln 2}-1=1.$$