8. Let $m=\sum_{k=1}^{2009} \frac{(k-1) \cdot 2^{k}}{(k-1)!}$, and define $[x]$ as the greatest integer not exceeding the real number $x$. Then $\left[m+\frac{1}{2}\right]=$
$\qquad$ .
8. 2 Detailed Explanation: $m=\sum_{k=1}^{2009} \frac{(k-1) \cdot 2^{k}}{(k-1)!}=\sum_{k=1}^{2009}\left(\frac{2^{k}}{k!}-\frac{2^{k+1}}{(k+1)!}\right)=\frac{2}{1}-\frac{2^{2010}}{2010!}=2-\frac{2^{2010}}{2010!}<2$, and $\frac{2^{2010}}{2010!}=\frac{2}{2010} \cdot \frac{2}{2009} \cdots \cdots \cdot \frac{2}{6} \cdot \frac{2^{5}}{5!}<\frac{2^{5}}{5!}<\frac{1}{2}$, so $2<m+\frac{1}{2}<\frac{5}{2}$, hence $\left[m+\frac{1}{2}\right]=2$.