this looks really simple but might end up quite complicated. i'm currently stuck though :/

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...

any partial ideas or references will be appreciated too!..