If $\sum_{n=1}^{\infty}\frac{a_n}{e^n}$ is convergent
denote$S_n=\sum_{k=1}^{n}a_n$
show that $\sum_{n=1}^{\infty}\frac{S_n}{e^n}$ is convergent.
If $\sum_{n=1}^{\infty}\frac{a_n}{e^n}$ is convergent
denote$S_n=\sum_{k=1}^{n}a_n$
show that $\sum_{n=1}^{\infty}\frac{S_n}{e^n}$ is convergent.
Use Abel's identity to show that:
$\sum_{n=1}^{k}\frac{S_n}{e^n}=1/(e^{-1}-1)[e^{-(k+1)}S_{k+1}-e^{-1}S_1-\sum_{n=1}^{k}\frac{a_{n+1}}{e^{n+1}}]$
Now the answer to your question follows very easily because the limit of the right hand side exists.