jmc

We can write $$\begin{array}{rl}\frac{k+2}{k!+(k+1)!+(k+2)!}& =\frac{k+2}{k![1+(k+1)+(k+1)(k+2)]}\\ & =\frac{k+2}{k!(k^2+4k+4)}\\ & =\frac{k+2}{k!(k+2{)}^2}\\ & =\frac{1}{k!(k+2)}\\ & =\frac{k+1}{k!(k+1)(k+2)}\\ & =\frac{k+1}{(k+2)!}.\end{array}$$Seeking a way to get the sum to telescope, we can express the numerator $k+1$ as $(k+2)-1.$ Then $$\frac{k+1}{(k+2)!}=\frac{(k+2)-1}{(k+2)!}=\frac{k+2}{(k+2)!}-\frac{1}{(k+2)!}=\frac{1}{(k+1)!}-\frac{1}{(k+2)!}.$$Therefore, $$\sum _{k=1}^{\infty }\frac{k+2}{k!+(k+1)!+(k+2)!}=\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\left(\frac{1}{4!}-\frac{1}{5!}\right)+\cdots =\boxed{\frac{1}{2}}.$$