I have the problem for computing the j-derivative of a logarithm, with $j\gg1$ \begin{equation} c_j=\frac{\partial^j}{\partial s^j}\log\left(1+Ae^s+Be^{2s}\right), \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=\frac{\partial^j}{\partial s^j}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}e^{ns}\left(A+Be^s\right)^n. \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

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