Summing Factorials

$\begin{aligned} S &=\dfrac{1 \cdot 1!}{(1+1!)(1+2!)}+\dfrac{2 \cdot 2!}{(1+2!)(1+3!)}+\dfrac{3 \cdot 3!}{(1+3!)(1+4!)}+ \cdots \text{2014 terms} \\ \\ &=\sum_{r=1}^{2014}\left[ \dfrac{r \cdot r!}{(1+r!)(1+(1+r)!)}\right] \end{aligned}$ Then write the $r$ in the numerator as $(r+1)!+1-(r+1)!-1+r.$ Doing so, we get $\begin{aligned} S&=\sum_{r=1}^{2014}\left[r! \cdot\dfrac{(r+1)!+1-(r+1)!-1+r}{(1+r!)(1+(1+r)!)}\right]\\ \\ &=\sum_{r=1}^{2014}\left[r!\cdot\dfrac{(r+1)!+r+1-(r+1)!-1}{(1+r!)(1+(1+r)!)}\right]\\ \\ &=\sum_{r=1}^{2014}\left[r! \cdot\dfrac{(r+1)(r!+1)-(r+1)!-1}{(1+r!)(1+(1+r)!)}\right]\\ \\ &=\sum_{r=1}^{2014}\left[r!\left( \dfrac{(r+1)}{1+(r+1)!}-\dfrac{1}{1+r!}\right)\right]\\ \\ &=\sum_{r=1}^{2014}\left[ \dfrac{(r+1)!}{1+(r+1)!}-\dfrac{r!}{1+r!}\right] \end{aligned}$

Now, let $f(n)=\dfrac{n!}{1+n!}$

So, $\begin{aligned} S&=\sum_{r=1}^{2014} \left[f(r+1)-f(r)\right]\\ \\ &=f(2)-f(1)+f(3)-f(2)+ \cdots +f(2015)-f(2014)\\ \\ &=f(2015)-f(1)\\ \\ &=\dfrac{2015!}{1+2015!}-\dfrac{1!}{1+1!}\\ \\ &= \boxed{\dfrac{2015!}{1+2015!}-\dfrac{1}{2}} \end{aligned}$

Hence, $\dfrac{A+B}{2C}=\dfrac{2015+1}{2\cdot 2}=\boxed{504}$

Note that k * k!/[(1+k!)(1+(k+1)!] = 1/(1+k!) - 1/[1+(k+1)!]. So

S

= 1/(1+1!)-1/(1+2!)+1/(1+2!)-1/(1+3!)+...+1/(1+2014!)-1/(1+2015!)

= 1/2-1/(1+2015!)

= (2015!-1)/[2(1+2015!)]

= 2015!/(1+2015!)-1/2.

Therefore A = 2015, B = 1, C = 2, and (A+B)/(2C) = 504.