How to evaluate $\sum_{r=1}^{r=15} \frac{r*2^r}{(r+2)!}$

Question

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.

Answer 1

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

$$\begin{align} \sum_{r=1}^{15}\frac{r2^r}{(r+2)!}&=\frac12\sum_{r=1}^{15} \left(\frac{2^{r+1}}{(r+1)!}-\frac{2^{r+2}}{(r+2)!}\right)\\\\ &=\frac{1}{2}\left(\frac{2^2}{(2)!}-\frac{2^{17}}{(17)!}\right)\\\\ &=1-\frac{2^{16}}{(17)!}\\\\ &\approx 0.999999999815748 \end{align}$$