How can I approximate $\sum\limits_{k=4}^{\infty}\Pr(X=k)[{\Pr(X\le k)}^6 - {\Pr(X\le k-4)}^6]$ for $\lambda \to +\infty$? $X$ is a Poisson random variable and the probability mass function is given by:
$$\Pr(X = k) = e^{-\lambda}\frac{{\lambda}^k}{k!}$$
I’ve got a probability function $f(\lambda)$
$$f(\lambda) = \sum\limits_{k=4}^{\infty}\Pr(X=k)[{\Pr(X\le k)}^6 - {\Pr(X\le k-4)}^6]$$
To date, I only find that ${\Pr(X\le k)}^6 - {\Pr(X\le k-4)}^6$ can be factorized as 
\begin{align*}
&{\Pr(X\le k)}^6 - {\Pr(X\le k-4)}^6 \\&= [\Pr(X=k)+ \Pr(X=k-1) + \Pr(X=k-2) + \Pr(X=k-3)]\cdot[ {\Pr(X\le k)}^5+{\Pr(X\le k)}^4{\Pr(X\le k-4)}+…+ {\Pr(X\le k-4)}^5]
\end{align*}
But I have no idea what to do next… Can I assume that $\Pr(X=k) \approx \Pr(X=k-1)$ if $\lambda \to +\infty$? Are there any better approximation for $f(\lambda)$? 
If there is a simple expression for $ f(\lambda) $ about $\lambda \to +\infty$ that would be best, but I’m open to whatever can be suggested. Thank you in advance!  
 A: By definition, $\sum\limits_kP(X=k)u(k)=E(u(X))$ for every bounded function $u$, hence, considering a Poisson $\lambda$ random variable $Y$ independent of $X$, one gets
$$
f(\lambda)=P(X\leqslant Y;Y\geqslant4)^6-P(X\leqslant Y-4)^6=(u(\lambda)-v(\lambda))^6-(u(\lambda)-w(\lambda))^6,
$$
with
$$
u(\lambda)=P(X\leqslant Y),\quad v(\lambda)=P(X\leqslant Y\leqslant3),\quad w(\lambda)=P(Y-3\leqslant X\leqslant Y).
$$
Since $Z_\lambda=(Y-X)/\sqrt{2\lambda}$ is approximately standard normal and
$$
u(\lambda)=P(Z_\lambda\geqslant0),\qquad w(\lambda)=P(0\leqslant Z_\lambda\leqslant3/\sqrt{2\lambda}),
$$
one sees that $u(\lambda)\to\frac12$ and $w(\lambda)\to0$. Likewise, $v(\lambda)\leqslant P(Y\leqslant3)\to0$ hence
$$
f(\lambda)=6\cdot\left(\tfrac12\right)^5(w(\lambda)-v(\lambda))+o(w(\lambda)+v(\lambda)).
$$
Now, $v(\lambda)\ll1/\sqrt\lambda$ and, if $(Z_\lambda)$ satisfies a local central limit theorem, the integer interval $[0,3]$ having length $4$, one might have $w(\lambda)\sim4/\sqrt{2\lambda}\cdot1/\sqrt{2\pi}$, hence
$$
f(\lambda)\sim6\cdot\left(\tfrac12\right)^5\cdot\frac2{\sqrt{\pi\lambda}}=\frac3{8\sqrt{\pi\lambda}}.
$$
This result is conditional on the (plausible) assertion that, for every fixed integer $k$, when $\lambda\to\infty$,
$$
P(Y-X=k)\sim\frac1{\sqrt{2\lambda}}\cdot\frac1{\sqrt{2\pi}}.
$$
Fortunately, asymptotics of Bessel functions of the first kind indicate that, for every $k\geqslant0$,
$$
P(Y-X=k)=\sum_{n\geqslant0}\mathrm e^{-2\lambda}\frac{\lambda^n}{n!}\frac{\lambda^{n+k}}{(n+k)!}=\mathrm e^{-2\lambda}\mathrm i^{-k}J_k(2\mathrm i\lambda)\sim\frac1{2\sqrt{\pi\lambda}}.
$$
Thus, the heuristics above holds and the result is proved.
