19. Let $a_{k}=\frac{k+1}{(k-1)!+k!+(k+1)!}$, then $a_{2}+a_{3}+\cdots+a_{1991}+\frac{1}{1992!}=$
19.  From $\frac{k+1}{(k-1)!+k!+(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}, \therefore \quad$ the original expression $=\frac{1}{2}$.