Prove that $\sum \limits_{n=0}^{\infty} \frac{n!}{(n+1)!+(n+2)!} = \frac{3}{4}$ I was playing around with factorials on Wolfram|Alpha, when I got this amazing result :
$$\sum \limits_{n=0}^{\infty} \dfrac{n!}{(n+1)!+(n+2)!} = \dfrac{3}{4}.$$
Evaluating the first few partial sums makes it obvious that the sum converges to $\approx 0.7$. But I am not able to prove this result algebraically. I tried manipulating the terms and introducing Gamma Function, but without success. 
Can anyone help me with this infinite sum ? Is there some well-known method of evaluating infinite sums similar to this ?
Any help will be gratefully acknowledged.
Thanks in advance ! :-)
EDIT : I realized that $(n!)$ can be cancelled out from the fraction and the limit of the remaining fraction as $n \to \infty$ can be calculated very easily to be equal to $0.75$. Very silly of me to ask such a question !!! Anyways you can check out the comment by @Did if this "Edit" section does not help.

Thanks to pjs36 and Did,
Notice that:
$$\begin{align}a_n&=\frac{n!}{(n+1)!+(n+2)!}\\&=\frac1{(n+1)+(n+1)(n+2)}\\&=\frac1{(n+1)(n+3)}\\&=\frac12\left(\frac1{n+1}-\frac1{n+3}\right)\end{align}$$
Thus, we get a telescoping series, leaving us with:
$$\sum_{n=0}^\infty\frac{n!}{(n+1)!+(n+2)!}=\frac12\left(\frac1{0+1}+\frac1{1+1}\right)=\frac34$$
