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

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

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Hint $\sum_{k=0}^\infty \frac{x^k}{k!}=e^x$ – leonbloy Feb 2 '13 at 15:41
Yeah I failed to realize that the problem was of that form of the Maclaurin series. – Brian Hauger Feb 2 '13 at 16:00
up vote 2 down vote accepted

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 $$\sum_{k=1}^\infty \frac{(\ln 2)^k}{k!}=e^{\ln 2}-1=1.$$

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