How to prove that $\sum_{k=0}^{n} \frac{k-1}{k!} =\frac{-n-1}{(n+1)!}$

Question

Just like in the question - how to prove that $\sum_{k=0}^{n} \frac{k-1}{k!} =\frac{-n-1}{(n+1)!}$ ? I will we thankful for any help.

Answer 1

$$\sum_{k=0}^{n}\frac{k-1}{k!}=\sum_{k=0}^{n}\frac{k}{k!}-\frac{1}{k!}$$ Now note that $\frac{k}{k!}=\frac{1}{(k-1)!}$ when $k\ge 1$. Therefore this sum can be written as $$\left(0-\frac{1}{0!}\right)+\left(\frac{1}{0!}-\frac{1}{1!}\right)+\ldots+\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)=\frac{-1}{n!}$$

Answer 2

Try this: \begin{align} \sum_{k=0}^n \frac{k-1}{k!} &= \sum_{k=0}^n \frac{k}{k!}-\sum_{k=0}^n\frac{1}{k!}\\ &= \sum_{k=1}^n \frac{1}{(k-1)!}-\sum_{k=0}^n\frac{1}{k!}\\ &= \sum_{k=0}^{n-1} \frac{1}{k!}-\sum_{k=0}^n\frac{1}{k!}\\ &= -\frac{1}{n!} \\ &= \frac{-(n+1)}{(n+1)!}. \end{align}