I have the problem for computing the j-derivative of a logarithm, with $j\gg1$
\begin{equation}
c\_j=\left.\frac{\partial^j}{\partial s^j}\log\left(1+Ae^s+Be^{2s}\right)\right|\_{s=0},
\end{equation}
being A and B real numbers ($0<A,B<1$). What I have done is to perform a Taylor expansion of the logarithm, finding
\begin{equation}
c\_j=\left.\frac{\partial^j}{\partial s^j}\sum\_{n=1}^\infty\frac{(-1)^{n+1}}{n}e^{ns}\left(A+Be^s\right)^n\right|\_{s=0}.
\end{equation}
Finally, I can use the Newton binomia expression and derive inside the sum
\begin{equation}
c\_j=\sum\_{n=1}^\infty\frac{(-1)^{n+1}}{n}\sum\_{k=0}^{n}{n\choose k}(2n-k)^jA^{n-k} B^{k}
\end{equation}
My question is to approximate this expression in the limit when $j\gg1$. I though about using some Stirling kind approximation, but I did not find any solution. Thanks for any help



aproximate sum involving binomial coefficients
Just a rough idea. Let $\alpha, \beta$ be the zeros of $1+Ax+Bx^2$, then for $j\geq 1$
$$c\_j = \left.\left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) )\right|\_{s=0} = \left.\left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) + \left(\frac{\partial}{\partial s}\right)^j \log(\beta-e^s)\right|\_{s=0}.$$


Now,
$$\log(\alpha-e^s) = \log \alpha + \log(1-\frac{e^s}{\alpha}) = \log \alpha - \sum\_{k=1}^{\infty} \frac{e^{ks}}{k\alpha^k}$$
and hence
$$ \left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) = - \sum\_{k=1}^{\infty} \frac{k^{j-1}e^{ks}}{\alpha^k} = - \frac{\sum\_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\left(\frac{e^s}{\alpha}\right)^{i+1}}{(1-\frac{e^s}{\alpha})^j},$$
and similarly for $\beta$. Here $\left\langle {\cdot\atop\cdot}\right\rangle$ are [Eulerian numbers](https://en.wikipedia.org/wiki/Eulerian_number).


Therefore,
$$c\_j = - \frac{\sum\_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\alpha^{j-1-i}}{(\alpha-1)^j} - \frac{\sum\_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\beta^{j-1-i}}{(\beta-1)^j}.$$


