Proving $(λ^d + (1-λ^d)e^{(d-1)s})^{\frac{1}{1-d}}\leq\sum\limits_{n=0}^\infty\frac1{n!}λ^{\frac{(d^n-1)d}{d-1}+n}s^ne^{-λs}$

$\def\e{\mathrm{e}} \def\veq{\mathrel{\phantom{=}}}$Denote $μ = λ^d$. It will be proved that\begin{align*} (μ + (1 - μ) \e^{(d - 1)s})^{-\frac{1}{d - 1}} \leqslant \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}} \end{align*} holds for $0 < μ < 1$, $s > 0$ and $d > 1$. Step 1 and Step 2 prepare for the proof of monotonicity in Step 3.

Lemma: For $0 < x <1$ and $r > 1$,$$ 1 - x^r \leqslant r(1 - x). $$ (This is trivial by considering the partial derivative of $x^r - rx$ with respect to $x$.)

Step 1: For $0 < μ < 1$, $d > 1$ and $n \in \mathbb{N}$,\begin{align*} &\veq \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k}))\\ &\geqslant μ \cdot (μ^{\frac{1}{d}} + d - 1)^n \cdot μ^{\frac{d^n - 1}{d - 1}} \geqslant μ \cdot μ^{\frac{n}{d}} d^n μ^{\frac{d^n - 1}{d - 1}} \geqslant μ \cdot μ^{\frac{n}{d}} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}}. \tag{1.1} \end{align*}

Proof: For $n = 0$, (1.1) becomes$$ 1 - μ^{\frac{1}{d}} (1 - μ) \geqslant μ \geqslant μ, $$ which is true because$$ 1 - μ^{\frac{1}{d}} (1 - μ) - μ = (1 - μ)(1 - μ^{\frac{1}{d}}) \geqslant 0. $$

For $n \geqslant 2$, since$$ (μ^{\frac{1}{d}} + d - 1)^n = \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k}, $$ then\begin{align*} &\veq \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ \cdot (μ^{\frac{1}{d}} + d - 1)^n μ^{\frac{d^n - 1}{d - 1}}\\ &= \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} \left( μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right). \tag{1.2} \end{align*}

For the $k = n$ term in (1.2), by the lemma,\begin{align*} &\veq μ^{\frac{n}{d}} \left( μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^n})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right) = μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ - μ^{\frac{1}{d}} (1 - μ^{d^n}))\\ &\geqslant μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ - μ^{\frac{1}{d}} d^n (1 - μ)) = μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ)(1 - μ^{\frac{1}{d}} d^n)\\ &\geqslant μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ)(1 - d^n) = - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ) \sum_{k = 0}^{n - 1} \binom{n}{k} (d - 1)^{n - k}, \end{align*} thus\begin{align*} (1.2) &\geqslant \sum_{k = 0}^{n - 1} \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} \left( μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right)\\ &\veq - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ) \sum_{k = 0}^{n - 1} \binom{n}{k} (d - 1)^{n - k}\\ &= \sum_{k = 0}^{n - 1} \binom{n}{k} (d - 1)^{n - k} \left( μ^{\frac{k}{d}} \left( μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right) - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ) \right). \tag{1.3} \end{align*}

For $0 \leqslant k \leqslant n - 1$, since $0 < μ < 1$ and $d > 1$, then $μ^{\frac{k}{d}}$, $μ^{\frac{d^k - 1}{d - 1}}$ and $μ^{d^k}$ are all decreasing with respect to $k$. Thus\begin{align*} &\veq μ^{\frac{k}{d}} \left( μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right) - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ)\\ &\geqslant μ^{\frac{n - 1}{d}} \left( μ^{\frac{d^{n - 1} - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^{n - 1}})) - μ^{\frac{d^n - 1}{d - 1}} \cdot μ \right) - μ^{\frac{n}{d}} μ^{\frac{d^n - 1}{d - 1}} (1 - μ)\\ &= μ^{\frac{n - 1}{d}} μ^{\frac{d^{n - 1} - 1}{d - 1}} \left( (1 - μ^{\frac{1}{d}} (1 - μ^{d^{n - 1}})) - μ^{d^{n - 1}} \cdot μ - μ^{\frac{1}{d}} μ^{d^{n - 1}} (1 - μ) \right)\\ &= μ^{\frac{n - 1}{d}} μ^{\frac{d^{n - 1} - 1}{d - 1}} \left( 1 - μ^{\frac{1}{d}} - μ^{d^{n - 1} + 1} + μ^{\frac{1}{d} + d^{n - 1} + 1}\right)\\ &= μ^{\frac{n - 1}{d}} μ^{\frac{d^{n - 1} - 1}{d - 1}} (1 - μ^{\frac{1}{d}})(1 - μ^{d^{n - 1} + 1}) \geqslant 0, \end{align*} which implies $(1.3) \geqslant 0$. Therefore,$$ \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^k})) \geqslant μ \cdot (μ^{\frac{1}{d}} + d - 1)^n μ^{\frac{d^n - 1}{d - 1}}. $$ Also, $0 < μ < 1$ and $d > 1$ implies $μ^{\frac{1}{d}} + d - 1 \geqslant d μ^{\frac{1}{d}}$, and by the lemma, $\displaystyle d^n \geqslant \frac{1 - μ^{d^n}}{1 - μ}$. Hence (1.1) holds for $0 < μ < 1$, $d > 1$ and $n \in \mathbb{N}$.

Step 2: For $0 < μ < 1$, $s > 0$ and $d > 1$,\begin{align*} &\veq \e^{(d - 1)s} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^n}))\\ &\geqslant μ \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}} \geqslant μ^{\frac{1}{d}} μ \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}}. \tag{2.1} \end{align*}

Proof: By Mertens' theorem,\begin{align*} &\veq \e^{(d - 1)s} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^n}))\\ &= \left( \sum_{n = 0}^\infty \frac{((d - 1)s)^n}{n!} \right) \left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^n})) \right)\\ &= \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{((d - 1)s)^{n - k}}{(n - k)!} \cdot \frac{(sμ^{\frac{1}{d}})^k}{k!} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^k}))\\ &= \sum_{n = 0}^\infty \frac{s^n}{n!} \sum_{k = 0}^n \binom{n}{k} μ^{\frac{k}{d}} (d - 1)^{n - k} μ^{\frac{d^k - 1}{d - 1}} (1 - μ^{\frac{1}{d}}(1 - μ^{d^k})). \tag{2.2} \end{align*}

By (1.1),\begin{align*} (2.2) &\geqslant \sum_{n = 0}^\infty \frac{s^n}{n!} μ \cdot μ^{\frac{n}{d}} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}} = μ \cdot \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}}\\ &\geqslant μ^{\frac{1}{d}} μ \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} \frac{1 - μ^{d^n}}{1 - μ} μ^{\frac{d^n - 1}{d - 1}}. \end{align*} Hence (2.2) holds for $0 < μ < 1$, $s > 0$ and $d > 1$.

Step 3: For $0 < μ < 1$ and $d > 1$,\begin{align*} f(s) &= \ln\left( \left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}} \right) (μ + (1 - μ) \e^{(d - 1)s})^{\frac{1}{d - 1}} \right)\\ &= \ln\left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \right) - sμ^{\frac{1}{d}} + \frac{1}{d - 1} \ln(μ + (1 - μ) \e^{(d - 1)s}) \end{align*} is increasing for $s > 0$.

Proof: Because for any $A > 0$, the series$$ \sum_{n = 0}^\infty \left( \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \right)' = \sum_{n = 1}^\infty \frac{n s^{n- 1} μ^{\frac{n}{d}}}{n!} μ^{\frac{d^n - 1}{d - 1}} = μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^{n + 1} - 1}{d - 1}} $$ converges uniformly for $s \in (0, A)$, then for any $s > 0$,\begin{align*} f'(s) &= \frac{\sum\limits_{n = 0}^\infty \left( \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \right)'}{\sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}}} - μ^{\frac{1}{d}} + \frac{1}{d - 1} \frac{(μ + (1 - μ) \e^{(d - 1)s})'}{μ + (1 - μ) \e^{(d - 1)s}}\\ &= \frac{μ^{\frac{1}{d}} \sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^{n + 1} - 1}{d - 1}}}{\sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}}} - μ^{\frac{1}{d}} + \frac{(1 - μ) \e^{(d - 1)s}}{μ + (1 - μ) \e^{(d - 1)s}}. \end{align*}

Define$$ A = μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} - μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^{n + 1} - 1}{d - 1}}, \quad B = \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}}. $$ Because\begin{align*} A &= μ^{\frac{1}{d}} \left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} - \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^{n + 1} - 1}{d - 1}} \right)\\ &= μ^{\frac{1}{d}} \left( \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} - \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \cdot μ^{d^n} \right)\\ &= μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{d^n}), \end{align*}\begin{align*} B - A &= \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} - μ^{\frac{1}{d}} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{d^n})\\ &= \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^n})), \end{align*} then\begin{align*} f'(s) \geqslant 0 &\Longleftrightarrow \frac{(1 - μ) \e^{(d - 1)s}}{μ + (1 - μ) \e^{(d - 1)s}} \geqslant \frac{A}{B}\\ &\Longleftrightarrow B(1 - μ) \e^{(d - 1)s} \geqslant A(μ + (1 - μ) \e^{(d - 1)s})\\ &\Longleftrightarrow (1 - μ)(B - A) \e^{(d - 1)s} \geqslant μA\\ &\Longleftrightarrow (B - A) \e^{(d - 1)s} \geqslant \frac{μ}{1 - μ} A\\ &\Longleftrightarrow \e^{(d - 1)s} \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} (1 - μ^{\frac{1}{d}} (1 - μ^{d^n})) \geqslant μ^{\frac{1}{d}} μ \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \frac{1 - μ^{d^n}}{1 - μ}, \end{align*} where the last inequality holds by (2.1). Hence $f(s)$ is increasing for $s > 0$.

Step 4: For $0 < μ < 1$, $s > 0$ and $d > 1$,\begin{align*} (μ + (1 - μ) \e^{(d - 1)s})^{-\frac{1}{d - 1}} \leqslant \sum_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}}. \tag{4.1} \end{align*}

Proof: From Step 3 it is known that $f(s)$ is increasing for $s > 0$, thus$$ \exp(f(s)) = \frac{\sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}}}{(μ + (1 - μ) \e^{(d - 1)s})^{-\frac{1}{d - 1}}} $$ is also increasing for $s > 0$. Because the series $\sum\limits_{n = 0}^\infty \frac{(sμ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-sμ^{\frac{1}{d}}}$ converges uniformly for $s \in [0, 1]$, then for any $s_0 > 0$,\begin{align*} &\veq \frac{\sum\limits_{n = 0}^\infty \frac{(s_0 μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s_0 μ^{\frac{1}{d}}}}{(μ + (1 - μ) \e^{(d - 1)s_0})^{-\frac{1}{d - 1}}} = \exp(f(s_0)) \geqslant \lim_{s \to 0^+} \exp(f(s))\\ &= \frac{\lim\limits_{s \to 0^+} \sum\limits_{n = 0}^\infty \frac{(s μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s μ^{\frac{1}{d}}}}{\lim\limits_{s \to 0^+} (μ + (1 - μ) \e^{(d - 1)s})^{-\frac{1}{d - 1}}} = \frac{\sum\limits_{n = 0}^\infty \lim\limits_{s \to 0^+} \frac{(s μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s μ^{\frac{1}{d}}}}{\left( μ + (1 - μ) \lim\limits_{s \to 0^+} \e^{(d - 1)s} \right)^{-\frac{1}{d - 1}}}\\ &= \frac{\lim\limits_{s \to 0^+} \e^{-s μ^{\frac{1}{d}}} + \sum\limits_{n = 1}^\infty \lim\limits_{s \to 0^+} \frac{(s μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s μ^{\frac{1}{d}}}}{\left( μ + (1 - μ) \lim\limits_{s \to 0^+} \e^{(d - 1)s} \right)^{-\frac{1}{d - 1}}} = \frac{1}{(μ + (1 - μ))^{-\frac{1}{d - 1}}} = 1, \end{align*} i.e.$$ \sum_{n = 0}^\infty \frac{(s_0 μ^{\frac{1}{d}})^n}{n!} μ^{\frac{d^n - 1}{d - 1}} \e^{-s_0 μ^{\frac{1}{d}}} \geqslant (μ + (1 - μ) \e^{(d - 1)s_0})^{-\frac{1}{d - 1}}. $$ Hence (4.1) holds for $0 < μ < 1$, $s > 0$ and $d > 1$.

Too long for a comment (but I can delete it if you want) so :

In fact the RHS is a polynomial more particulary a Jensen polynomial :

Why? Because the coefficients fulfill the conditions also called Turan's inequality: $$\sigma_k^2\geq \sigma_{k-1}\sigma_{k+1}$$

Proof (here $\lambda=\sigma$): We have to prove :

$$\sigma^{2\frac{d^n-1}{d-1}+2n}\geq \sigma^{\frac{d^{n-1}-1}{d-1}+n-1}\sigma^{\frac{d^{n+1}-1}{d-1}+n+1}$$

After simplification we have :

$$\sigma^{2\frac{d^n-1}{d-1}}\geq\sigma^{\frac{d^{n-1}-1}{d-1}}\sigma^{\frac{d^{n+1}-1}{d-1}}$$ We take the logarithm on each side we get : $${2\frac{d^n-1}{d-1}}ln(\sigma)\geq\frac{d^{n-1}-1}{d-1}ln(\sigma)+\frac{d^{n+1}-1}{d-1}ln(\sigma)$$

The logaritm is negative so we reverse the inequality :

$${2(\frac{d^n-1}{d-1})}\leq\frac{d^{n-1}-1}{d-1}+\frac{d^{n+1}-1}{d-1}$$

Finally we get :

$$2\leq \frac{1}{d}+d$$ Wich is obvious

So it's a serious way to prove it but now I have not the time to conclude

Edit : The associated polynomials to Jensen polynomials are :

$$g_n(t)=\sum_{k=0}^{n}{n\choose k}\sigma_tx^t$$

Futhermore we know (let $g_n(t)=g_n(t)(\phi;x)$) then the polynomials $g_n(t)$ are generated by : $$e^t\phi(xt)=\sum_{k=0}^{\infty}g_n(x)\frac{t^n}{n!}$$ Theorem A from this link asserts this : the function $\phi(x)$ belongs to L-P if and only if all the polynomials $g_n(\phi;x)$,$n=1,2,\cdots$ have only real zeros .

From this link we can say that the zeros of Jensen polynomials are simple and negative so we can express the Jensen polynomials like this :

$$f(z)=cz^me^{-\alpha z^2+\beta z }\prod_{n=0}^{\infty}(1-\frac{z}{z_n})e^{\frac{z}{z_n}}$$

So the infinite sum becomes a product... maybe it could help .