# Solving $\sum_{\alpha=1}^{\mu}\frac{1}{\alpha!}\Big(1-\alpha(2 \mu- \alpha)\Big(\frac{M-2l}{2 \mu^2 KM}\Big)\Big)^{\lambda}$

I asked about a part of this problem some time ago, and Didier Piau 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 on $(1-\frac{\alpha(2 \mu- \alpha)(M-2l)}{2 \mu^2 KM})^{\lambda} \leq e^{-\frac{alpha(2 \mu-\alpha)(M-2l) \lambda}{2 \mu^2 KM}}$. 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
$\frac{1}{\sqrt{2 \pi}}\sum_{\alpha=1}^{\mu}e^{-\alpha(2 \mu-\alpha)z+\alpha +log \alpha^{-(\alpha+.5)}}$. For sufficiently large $\mu$ this expression can be approximated with Riemannian sums. To avoid 0, the bounds of the integral become $[1,2]$ and the approximation of this sum is $\frac{\mu}{\sqrt{2 \pi}}\int_{1}^{2}e^{-\mu \alpha(2 \mu -\mu \alpha)z+\mu \alpha+log(\mu \alpha)^{-(\mu \alpha+.5)}}$ 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.

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I don't understand the justification for setting $l=0$ and $\lambda=\mu$. Can we set $l$ and $\lambda$ to whatever we want? I mean, if you try $l=M/2$, for example... (Edit: also, the title is misleading if you're only searching for a nice approximation instead of necessarily a closed-form solution.) –  anon Jul 19 '11 at 7:51
What are you trying to do with the sum? Get bounds? Approximate it? Get asymptotical behavior? Be precise on what you want. –  Patrick Da Silva Jul 19 '11 at 7:57
yes, an approximation will do. $l=0$ and $\lambda=\mu$ should hold. –  sigma.z.1980 Jul 19 '11 at 7:58

Similar to your previous post and as explained in the comments of the present post by others, some crucial information is missing. Let us however assume that $c=(M-2l)/(2KM)$ and $\lambda$ are fixed, that $c$ is nonnegative and that $\mu\to+\infty$. Then one considers $$S_\mu=\sum_{a=1}^{+\infty}x_\mu(a),\quad x_\mu(a)=\frac1{a!}\left(1-ca(2\mu-a)/\mu^2\right)^\lambda\mathbf{1}_{a\le\mu}.$$ For every fixed $a$, $x_\mu(a)\to x(a)=\displaystyle\frac1{a!}$ when $\mu\to+\infty$ with $x_\mu(a)\le x(a)$ and $S=\displaystyle\sum_{a=1}^{+\infty}x(a)$ converges, hence Lebesgue convergence theorem shows that $S_\mu\to S=\mathrm{e}-1$.
Edit If, on the contrary, one assumes that $c$ is nonnegative and fixed and that $\lambda=\mu$ with $\mu\to+\infty$, a similar argument yields the limit $\exp(\mathrm{e}^{-2c})-1$.
The asymptotic approach is 'the last resort'. I'd like to solve it assuming 'relatively small' $\mu$. –  sigma.z.1980 Jul 19 '11 at 21:24