I am interested in the asymptotic behavior as $M\rightarrow\infty$ of the following function: $$ g_{M}(x)= e^{-x} \sum_{k=0}^{M} \frac{\left(x\left(1-\frac{\lambda}{M}\right)\right)^{k}}{k!},$$ where $\lambda>0$ is a constant and $x>0$.
I derived that $$g_{M}(x)\sim \frac{1}{2}e^{-\frac{\lambda}{M}x}\left[1 + \mathrm{erf}\left(\frac{M-x\left(1-\frac{\lambda}{M}\right)}{\sqrt{2x\left(1-\frac{\lambda}{M}\right)}}\right)\right], \quad \mathrm{as} \quad M \to \infty, $$ where $\mathrm{erf}(z)$ is the error function.
Outline of my approach: I believe this method is known as Laplace's method or saddle-point approximation for sums.
For large $M$, the sum is to be dominated by terms near its peak, which occurs around $k \approx \mu$, where $ \mu = x \left(1 - \frac{\lambda}{M}\right). $
This sum resembles the Poisson distribution. Its variance $\sigma^2$ is approximately $ \mu $, leading me to approximate the sum using Stirling's approximation and Gaussian distribution methods.
The Gaussian approximation is then: $$ \sum_{k=0}^{M} \frac{\left(x\left(1 - \frac{\lambda}{M}\right)\right)^k}{k!} \approx \frac{1}{\sqrt{2 \pi \sigma^2}} \int_0^{M} \exp\left(-\frac{(k - \mu)^2}{2 \sigma^2}\right) \, dk. $$ This integral corresponds to the cumulative distribution function (CDF) of a normal distribution, which can be expressed using the error function, yielding the result, which I verified numerically.
My question:
Is there a simpler or more direct method to derive this result, or perhaps a chance at obtaining an elegant closed-form expression for $g_M(x)$ (for finite $M$)? I would appreciate any insights or suggestions.
