Feller's limit $\lim_{n\to\infty}e^{-n}\sum_{k=0}^{an}\frac{n^k}{k!}$ In the second volume of Feller's "An Introduction to Probability Theory and Its Applications", section VII.6 on Inversion formulas for Laplace transform starts with the limit
$\displaystyle{\lim_{\lambda\to\infty}e^{-\lambda\theta}\sum_{k\leq\lambda x}\frac{(\lambda\theta)^k}{k!}}$
or, in another form, 
$\displaystyle{\lim_{n\to\infty}e^{-n}\sum_{k=0}^{an}\frac{n^k}{k!}}$.
Feller says that the above limit is $0$ if $a<1$ and $1$ if $a>1$, and refers to previous chapters.
I am interested in direct computation of the limit (related question here does not seen illuminating in this aspect), without references to Central Limit theorem (like seen in other similar questions; usage of Poisson variables is possible).
 A: This is in a chapter called "Laws of Large Numbers. Applications in Analysis"  
The CLT has very little to do with this problem.  On the other hand the Strong (and Weak) Law of Large numbers has a lot to do with the limit.  
What Feller has shown is an example of a 0-1 law.  I'd suggest considering increasing sequences and positive integer valued $\lambda$.  
You may interpret $X_{\lambda}$ as the sum of $\lambda$ iid poisson random variables, each having mean $\theta$, hence $X_{\lambda}$  is poisson with mean $\lambda \theta$.    
Feller then says that 
the probability of the event
$\big \vert X_{\lambda} - \lambda \theta \big \vert \gt \lambda \epsilon $  is small for large enough $\lambda$ 
i.e.
$P\Big(\big \vert X_{\lambda} - \lambda \theta \big \vert \gt \lambda \epsilon\Big) \lt \epsilon' $
for any $\epsilon \gt 0$ and $\epsilon' \gt 0$ by selecting large enough $\lambda$ 
a more conventional form for this is 
$P\Big(\big \vert \frac{X_{\lambda}}{\lambda} - \theta \big \vert \gt \epsilon\Big) \lt \epsilon' $
which is something the SLLN tells you.  (You could actually use WLLN here if you wanted.)  
crude note: this implies that $P\Big(\big \vert X_{\lambda} - \lambda \theta \big \vert = \lambda \epsilon\Big)$ may be made arbitrarily small
But for any positive $x \lt  \theta$
$e^{-\lambda \theta}\sum_{k\leq \lambda x} \frac{(\lambda \theta)^k}{k!} $
$= P\big(X_{\lambda}\leq \lambda x\big) $
$\leq P\Big( X_{\lambda} \leq \lambda (\theta -\epsilon)\Big) $  (by selecting small enough $\epsilon$, making use of monotone non-decreasing nature of CDFs)
$= P\Big(\lambda\epsilon  \leq \lambda \theta - X_{\lambda})\Big) $
$= P\Big(\lambda \theta - X_{\lambda}\geq \lambda\epsilon \Big) $
$\leq P\Big(\lambda \theta - X_{\lambda}\geq \lambda\epsilon \Big) + P\Big(X_{\lambda} -\lambda \theta \geq \lambda\epsilon \Big) $
$= P\Big(\big \vert X_{\lambda} - \lambda \theta \big \vert \gt \lambda \epsilon\Big) + P\Big(\big \vert X_{\lambda} - \lambda \theta \big \vert = \lambda \epsilon\Big) $
$ \lt \frac{\epsilon'}{2} + \frac{\epsilon'}{2} $
$= \epsilon'$
where of course probabilities are bounded below by zero, and in this case bounded above by $\epsilon'$  which may be made arbitrarily small by selecting sufficiently large $\lambda$.  This gives the zero, whenever $x \lt \theta$ 
A near identical argument on the complement gives the 1 when $\theta \lt x$ 
This sequential result then implies the desired result on tthe next page (233 of the second edition) where Feller shows, in effect, the injectivity of the laplace transform of a real non-negative random variables' distribution (with the usual caveats: assuming the transform exists, and equivalence up to a set of measure zero)  
