Explain why $\sum\limits_{t=1}^{\infty}\frac{e^{-1}}{t!}=1 - \frac{1}{e}$

Question

Just want to know the step in between calculating this that I forgot about when I took probability theory 5 years ago. We have $$E\left(\frac{1}{1+X}\right) = \sum_{k=0}^{\infty}\frac{1}{k+1}P(X=k) = \sum_{k=0}^{\infty}\frac{e^{-1}}{(k+1)!} = \sum_{t=1}^{\infty}\frac{e^{-1}}{t!}$$

This is suppose to equal $1 - \frac{1}{e}$ but I seem to have forgotten the step after this. Any help is greatly appreciated.

Answer 1

@Wolfi, Here is the missing step in detail:

$$\sum\limits_{t=1}^{\infty}\frac{e^{-1}}{t!}=e^{-1}\left(\sum\limits_{t=1}^{\infty}\frac{1}{t!}\right)=e^{-1}\left(\left(\sum\limits_{t=0}^{\infty}\frac{1}{t!}\right)-1\right)=e^{-1}(e-1)=1 - \frac{1}{e}$$

Answer 2

The sum on the right is equal to $e^{-1}(e-1)=1-\frac{1}{e}$ since $e=\sum_{t=0}^{\infty}\frac{1}{t!}$.