Solving $\sum\limits_{\alpha=1}^{\mu}\frac{1}{\alpha!}\left(1-\alpha(2 \mu- \alpha)\left(\frac{M-2l}{2 \mu^2 KM}\right)\right)^{\lambda}$
I asked about a part of this problem some time ago, and user @Did gave some suggestions, but I would like nevertheless to ask it again, since I could not arrive at a sensible solution.
I use Stirling approximation $\frac{1}{\alpha!} \approx (\frac{e}{\alpha})^{\alpha}\frac{1}{\sqrt{2 \pi \alpha}}$ and the upper bound $\left(1-\frac{\alpha(2 \mu- \alpha)(M-2l)}{2 \mu^2 KM}\right)^{\lambda} \leq \exp\left(-\frac{\alpha(2 \mu-\alpha)(M-2l) \lambda}{2 \mu^2 KM}\right)$ Setting $l=0$ and $\lambda=\mu$ the expression under the exponential function simplifies to $-\frac{\alpha(2 \mu-\alpha)}{2 \mu K}$ so the expression (given that $\alpha^{-\alpha}=e^{log\alpha^{-\alpha}}$ and setting for simplicity $z=\frac{1}{2 \mu K}$) becomes
$\frac{1}{\sqrt{2 \pi}}\sum_{\alpha=1}^{\mu}\exp\left(-\alpha(2 \mu-\alpha)z+\alpha +\log \alpha^{-(\alpha+.5)}\right)$ For sufficiently large $\mu$ this expression can be approximated with Riemann sums. To avoid 0, the bounds of the integral become $[1,2]$ and the approximation of this sum is $\frac{\mu}{\sqrt{2 \pi}}\int_{1}^{2}\exp(-\mu \alpha(2 \mu -\mu \alpha)z+\mu \alpha+\log(\mu \alpha)^{-(\mu \alpha+.5)})d\alpha$ I checked this expression nmerically though, for different values of $\mu$, and it doesn't give a good approximation at all. I also tried expanding in Taylor series without much success. Did I make a mistake somewhere? Any suggestions are massively welcome.