how do i find the sum to infinity of the sum of (2r+1)/r! where r>=1. i have no clue about sums and factorials.
$\displaystyle \sum_{r = 1}^{\infty} \frac{1}{r!} = e-1$
$\displaystyle 2 \sum_{r = 1}^{\infty} \frac{r}{r!} = 2 \sum_{r = 1}^{\infty} \frac{1}{(r-1)!} = \sum_{r = 0}^{\infty} \frac{1}{r!} = e$ (simply shifted the "index" by 1)
Adding, we get
$\displaystyle \sum_{r = 1}^{\infty} \frac{2r+1}{r!} = 2e + (e-1) = 3e - 1$