Using Stirling's formula, show that, if $\lambda \rightarrow \infty$, then for fixed $\alpha < \beta$ \begin{equation} \sum_{\lambda + \alpha \sqrt{\lambda}< k < \lambda + \beta \sqrt{\lambda}}p(k; \lambda) \rightarrow \mathcal{R}(\beta) - \mathcal{R}(\alpha) \end{equation}
where $\mathcal{R}(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-y^2/2}dy$
This is what I have thus far: By definition we have $p(k; \lambda) = e^{-\lambda} \frac{\lambda^k}{k!}\sim e^{-\lambda} \frac{\lambda^k}{\sqrt{2\pi k}\left(\frac{k}{e}\right)^k}$ using $\lambda \rightarrow \infty$ and Stirling's Approximation. We note that when $\lambda \rightarrow \infty$, we must have $e^{-\lambda} \sim 1 + \frac{1}{\lambda}$. Thus, $p(k; \lambda) =\left(1 + \frac{1}{\lambda}\right) \frac{\lambda^k}{\sqrt{2\pi k}\left(\frac{k}{e}\right)^k}$.
I'm not sure if I'm walking down a dead end, and would appreciate some feedback.
To prove$$\lim_{\lambda\to\inf}\left(\sum_{\frac{k-\lambda}{\sqrt{\lambda}}\in[\alpha,\,\beta],\,k\in\Bbb Z^\ast}\frac{\lambda^ke^{-\lambda}}{k!}\right)=\int_\alpha^\beta\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy,$$we use a discrete version of Laplace's method, together with the limit-of-sums definition of a Riemann integral. Since$$-\ln p(k)=\lambda-k\ln \lambda+\ln k!\in\lambda-k\ln\lambda+(k+\tfrac12)\ln k-k+\ln\sqrt{2\pi}+O(\tfrac1k),$$the first two derivatives thereof with respect to $k$, treated as if continuous, are$$-\ln\lambda+\ln k+\tfrac{1}{2k}+O(\tfrac{1}{k^2}),\,\tfrac1k-\tfrac{1}{2k^2}+O\left(\tfrac{1}{k^3}\right).$$To a first-order approximation, $-\ln p$ is minimized at $k\approx\lambda$ with second derivative $\approx\tfrac{1}{\lambda}$, so$$-\ln p\approx-\ln p(\lambda)+\tfrac{(k-\lambda)^2}{2\lambda}.$$This approximates the distribution as Gaussian, so unitarity fixes the proportionality constant. With $y:=\tfrac{k-\lambda}{\sqrt{\lambda}}\approx N(0,\,1)$, the sum operator $\sum_{\alpha\le y\le\beta}$ is applied to approximately $\frac{1}{\sqrt{2\pi}}e^{-y^2/2}$, giving the desired result.