Asymptotic Expansion (Taylor) when n goes to infinity of $ e^{-an}\sum_{k=n+1}^{\infty} \frac{(na)^k}{k!} $

Question

I'm seeking theses asymptotic expansions :

$$ e^{-an}\sum_{k=n+1}^{\infty} \frac{(na)^k}{k!} $$

$$ e^{-an}\sum_{k=1}^{n-1} \frac{(na)^k}{k!} $$

in terms of n going to the infinity.

Thanks you for your help.

Answer 1

I assume $a>0$.

Notice that your tail sum is the probability that a Poisson($an$) variable exceeds $n$. This is the probability that the average of $n$ Poisson($a$) variables exceeds $1$. For $a \neq 1$ the result of the limit comes from the weak law of large numbers: if $a<1$ then the result is $0$, if $a>1$ then the result is $1$. For $a=1$ you can use the central limit theorem: the average behaves asymptotically like $N(1,1/n)$ so the limit is $1/2$, from symmetry of the normal distribution (even though the underlying Poisson distribution isn't symmetric). This is highly non-obvious from the analysis point of view.

Since the Poisson distribution has all cumulants, you can try to use Edgeworth series to get higher order asymptotics.