Expectation Poisson Distribution

A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and $10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. What is the expected amount paid to the company under this policy during a one-year period?

I know how to calculate the expectation and what the series is. I'm having problems with the summations. I know it should involve:

$$\sum_{k=2}^{+\infty} \frac{(1.5)^k}{k!}$$

Answer

Let $X$ be the number of snowstorms occurring in the given year and let $Y$ be the amount paid to the company. Call one unit of money $\$ 10{,}000$.

Then $Y$ takes the value $0$ when $X=0$ or $X=1$, the value $1$ when $X=2$, the value $2$ when $X=3$, etc..

The expected payment is
$$\eqalign{ \Bbb E(Y) &=\sum_{k=2}^\infty (k-1)P[X=k]\cr &=\sum_{k=2}^\infty (k-1) e^{-1.5}{(1.5)^k\over k!}\cr &=\sum_{k=1}^\infty (k-1) e^{-1.5}{(1.5)^k\over k!}\cr &= \sum_{k=1}^\infty k e^{-1.5}{(1.5)^k\over k!} -\sum_{k=1}^\infty e^{-1.5}{(1.5)^k\over k!}\cr &=\underbrace{ \sum_{k=0}^\infty k e^{-1.5}{(1.5)^k\over k!}}_{\text{mean of } X} - \biggl(-e^{-1.5}+\underbrace{\sum_{k=0}^\infty e^{-1.5}{(1.5)^k\over k!}}_{=1}\biggr)\cr &=1.5+e^{-1.5}- 1\cr &=0.5+e^{-1.5}\cr &\approx .7231\,\text{units}. }$$