Hello it's related to this $(\lambda^d + (1-\lambda^d) e^{(d-1)s})^{\frac{1}{1-d}} \leq \sum_{n=0}^\infty \frac{\lambda^{\frac{(d^n-1)d}{d-1}+n}s^n}{n!} e^{-\lambda s}$ it was :

Let $\lambda \in (0,1), s \in (0,\infty), d \in \{2,3,\dots\}$ and show that in this case the following inequality holds: $$(\lambda^d + (1-\lambda^d) e^{(d-1)s})^{\frac{1}{1-d}} \leq \sum_{n=0}^\infty \frac{\lambda^{\frac{(d^n-1)d}{d-1}+n}s^n}{n!} e^{-\lambda s},$$

I have found (numericaly speaking) the following refinement :

$$(\lambda^d + (1-\lambda^d) e^{(d-1)s})^{\frac{1}{1-d}} \le \frac{1}{\Gamma(e^{(-0.9s)})}\leq \sum_{n=0}^\infty \frac{\lambda^{\frac{(d^n-1)d}{d-1}+n}s^n}{n!} e^{-\lambda s},$$

Suprisingly it not depends on the factors $\lambda$ and $d$ and it involves the reciprocal of the Gamma function .

So my question is how to prove it ?

I will post my try later (I have to work for my job) but it's based on the Weierstrass theorem of factorization .

Thanks a lot .

Edit : We have $\lambda \in (0;0.9]$ otherwise it doesn't work as precised in the comment .

  • $\begingroup$ See my edits It works normally... $\endgroup$ – user448747 Feb 17 '18 at 17:20
  • $\begingroup$ And if we restrict the value of $\lambda$ to $[a;b]$ with $b\leq 0.5$ it works ? $\endgroup$ – user448747 Feb 18 '18 at 11:21

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