How can one show that $\sum_{n=0}^\infty\frac{n}{n!}=e$?
I understand that $$ \sum_{n=0}^\infty\frac{x^n}{n!}=e^x $$ and that letting $x=1 $would give $$ \sum_{n=0}^\infty\frac{1}{n!}=e $$
But why does the sum $\sum_{n=0}^\infty\frac{n}{n!}$ give an answer of $e$ also?