Where does it converges?

Question

Where does $\lim_{n\rightarrow\infty}\frac{\sum_{k=n}^{\infty}\frac{\lambda^{k}}{k!}e^{-\lambda}}{\sum_{k=n-1}^{\infty}\frac{\lambda^{k}}{k!}e^{-\lambda}}$ converges? I got the following, but I don't know how to continue. I'm confused whether it is $0$ or $1?$ $$ \lim_{n\rightarrow\infty}\frac{\sum_{k=n}^{\infty}\frac{\lambda^{k}}{k!}e^{-\lambda}}{\sum_{k=n-1}^{\infty}\frac{\lambda^{k}}{k!}e^{-\lambda}}=\lim_{n\rightarrow\infty}\frac{1-\sum_{k=0}^{n-1}\frac{\lambda^{k}}{k!}e^{-\lambda}}{1-\sum_{k=0}^{n-2}\frac{\lambda^{k}}{k!}e^{-\lambda}}=\lim_{n\rightarrow\infty}\frac{1-\sum_{k=0}^{n-2}\frac{\lambda^{k}}{k!}e^{-\lambda}-\frac{\lambda^{n-1}}{(n-1)!}e^{-\lambda}}{1-\sum_{k=0}^{n-2}\frac{\lambda^{k}}{k!}e^{-\lambda}}=$$ $$=\lim_{n\rightarrow\infty}1-\frac{\frac{\lambda^{n-1}}{(n-1)!}e^{-\lambda}}{1-\sum_{k=0}^{n-2}\frac{\lambda^{k}}{k!}e^{-\lambda}}=... $$

Answer 1

$\sum_{k \geq n}\frac{\lambda^k}{k!}\sim_{+\infty} \frac{\lambda^n}{n!}$ and so the limit will be $0$