Summation of a Series!

Before we start to tackle the sum, let's focus on the expression that we are going to sum. Notice that $\begin{aligned} \frac{k^2-3}{(k+1)!}&=&\frac {k^2+2k+1-2k-4}{(k+1)!}\\ &=&\frac {1}{(k-1)!}-\frac{1}{k!}-\frac {2}{(k+1)!} \end{aligned}$ We can now rewrite the summation as $\begin{aligned} S&=&\sum^{2006}_{k=1}(-1)^k\frac{k^2-3}{(k+1)!}\\&=&\sum^{2006}_{k=1}(-1)^k\frac {1}{(k-1)!}-\sum^{2006}_{k=1}(-1)^k\frac {1}{k!}-\sum^{2006}_{k=1}(-1)^k\frac {2}{(k+1)!}\\&=&-1/0!+1/1!-1/2!+1/3!-1/4+...+1/2005!\\&&+1/1!-1/2!+1/3!-1/4!+...+1/2005!-1/2006!\\&&+2/2!-2/3!+2/4!-...-2/2005!+2/2006!-2/2007!\\&=&1+1/2006!-2/2007! \end{aligned}$ Therefore, $S-1=1/2006!-1/2007!=2005/2007!$ , which gives us the final answer as $A+B=2005+2007=4012.$