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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.

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  • $\begingroup$ 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.) $\endgroup$ – anon Jul 19 '11 at 7:51
  • $\begingroup$ What are you trying to do with the sum? Get bounds? Approximate it? Get asymptotical behavior? Be precise on what you want. $\endgroup$ – Patrick Da Silva Jul 19 '11 at 7:57
  • $\begingroup$ yes, an approximation will do. $l=0$ and $\lambda=\mu$ should hold. $\endgroup$ – sigma.z.1980 Jul 19 '11 at 7:58
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Similar to your previous post, and, as others explained in the comments to the present post, some crucial information is missing from the question.

Let us however assume that $c=(2KM)^{-1}(M-2l)$ and $\lambda$ are fixed, that $c\geqslant0$ and that $\mu\to+\infty$. Then one considers $$ S_\mu=\sum_{a=1}^{+\infty}\frac1{a!}x_\mu(a)$$ with $$ x_\mu(a)=\left(1-ca\mu^{-2}(2\mu-a)\right)^\lambda\mathbf{1}_{a\le\mu} $$ Thus, for every fixed $a$, $x_\mu(a)\to1$ when $\mu\to+\infty$ with $x_\mu(a)\leqslant1$, and the series $\sum\frac1{a!}$ converges, hence, by Lebesgue convergence theorem, $$\lim_{\mu\to\infty}S_\mu=\sum_{a=1}^{+\infty}\frac1{a!}=\mathrm{e}-1$$ If, on the contrary, one assumes that $c\geqslant0$ is fixed and that $\lambda=\mu$ with $\mu\to+\infty$, then a similar argument yields the limit $$\lim_{\mu\to\infty}S_\mu=\exp(\mathrm{e}^{-2c})-1$$

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  • $\begingroup$ If the asymptotic regime considered in my answer is not the one you are interested in, you should definitely explain which one you consider. $\endgroup$ – Did Jul 19 '11 at 13:01
  • $\begingroup$ The asymptotic approach is 'the last resort'. I'd like to solve it assuming 'relatively small' $\mu$. $\endgroup$ – sigma.z.1980 Jul 19 '11 at 21:24

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