How to evaluate $\sum_{r=1}^{r=15} \frac{r*2^r}{(r+2)!}$ How can one evaluate the sum
$$\sum_{r=1}^{r=15} \frac{r\,2^r}{(r+2)!}$$
I am not able to start because of the factorial in the denominator.
I have tried decomposing into partial fractions, but have failed.

Note that we can write 
$$\begin{align}
\frac{r}{(r+2)!}&=\frac{(r+2)-2}{(r+2)!}\\\\
&=\frac{1}{(r+1)!}-\frac{2}{(r+2)!}
\end{align}$$
Therefore, we have
$$\begin{align}
\sum_{r=1}^{15}\frac{r2^r}{(r+2)!}&=\frac12\sum_{r=1}^{15} \left(\frac{2^{r+1}}{(r+1)!}-\frac{2^{r+2}}{(r+2)!}\right)\\\\
&=\frac{1}{2}\left(\frac{2^2}{(2)!}-\frac{2^{17}}{(17)!}\right)\\\\
&=1-\frac{2^{16}}{(17)!}\\\\
&\approx 0.999999999815748
\end{align}$$
