Series equal to e

Question

I'm having trouble convincing myself why

$$\sum_{k = 0}^{\infty} \frac{k}{k!} = e.$$

As I was under the impression that only

$$\sum_{k = 0}^\infty \frac{1}{k!} = e$$

by definition.

By writing out terms of the first series

$$\sum_{k = 0}^{\infty} \frac{k}{k!} = 0 + 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ...$$

Shouldn't it then be $1 + e$, not just $e$?

Thanks!

Answer 1

Since $\frac 0{0!}=0$, we have $$\sum_{k=0}^\infty\frac k{k!}= \sum_{k=1}^\infty\frac k{k!}=\sum_{k=0}^\infty\frac {k+1}{(k+1)!}=\sum_{k=0}^\infty\frac 1{k!}=e.$$