# exponential type series question

suppose that for every k, $\sum_{n=0}^\infty \frac{a_{k+n}}{n!}=e$. does this imply that $a_k \to 1$?
the above condition can also be rewritten as $\sum_{n=0}^\infty \frac{1-a_{k+n}}{n!}=0$ of course, which makes it even more painfully obvious, but I still can't prove this...
I need also to figure out a more general question: does $\lim_{k\to\infty}\sum_{n=0}^\infty \frac{a_{k+n}}{n!}=e$ imply that $a_k \to 1$, but clearly knowing the answer for the initial question would be a good start...