TITLE: Showing that a family of polynomials has positive and real roots.
QUESTION [13 upvotes]: Hi everybody, for my research I am dealing with the following function: 
$$\alpha_n(x):=\left.\frac{\partial^{2n+1}}{\partial z^{2n+1}}\frac{\sinh(z)}{\cosh(z)-1+x}\right|_{z=0},\quad n\in \mathbb{N},$$
It is possible to show that
$$\alpha_n(x)=\frac{P_n(x)}{x^{n+1}},$$ where $P_n(\cdot)$ is a polynomial of order $n$ in $x$, having integer coeffients.
To make few concrete examples
$$\alpha_0(x)=\frac{1}{x}$$ 
$$\alpha_1(x)=\frac{-3+x}{x^2}$$ 
$$\alpha_2(x)=\frac{30-15 x+x^2}{x^3}$$ 
$$\alpha_3(x)=\frac{-630+420 x-63 x^2+x^3}{x^4}$$ 
$$\alpha_4(x)=\frac{22680-18900 x+4410 x^2-255 x^3+x^4}{x^5}$$ 
and so on.
What I would need to show (and it is veryfied for all the special cases I was able to compute, like those above) is that all the roots of $P_n(x)$ (and therefore those of $\alpha_n(x)$) are real and strictly greater than 2. 
An explicit albeit complicated expression for $\alpha_n(x)$ can be obtained, namely:
$$\begin{equation*}
\begin{array}{ll}
\alpha_n(x)=&x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j+1}^{n} (2k)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k}\sum_{i=j}^{k}\left[ {i+1\choose j+1}\binom{2k}{2 i+1}-{i\choose j+1}\binom{2k}{2 i}\right](-2)^{j+1-2k}+\\
&x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j}^{n}{(2k+1)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k+1}}\sum_{i=j}^{k} \left[ {i+1\choose j+1}\binom{2k+1}{2 i+1}-{i\choose j+1}\binom{2k+1}{2 i}\right](-2)^{j-2k},
\end{array}
\end{equation*}$$
where the number between  the curly brakets are the Stirling number of the second kind; Moreover, 
$$\alpha_n(2)=-\sum_{k=1}^{2n+1}{k!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{k}}(-2)^{-k}\neq0.$$
If someone is interested, I can post more on how I got these expressions.
Thanks in advance to everybody that will try to help me.
Best Regards 
Enzo

REPLY [4 votes]: This paper by P. Bränden and this paper by M. Visontai & N. Williams give a somewhat general approach to proving real rootedness of polynomials (especially those coming up combinatorially).