limit of sum with binomial coefficients

Question

I have the problem to compute next double sum \begin{equation} \sum_{n=2}^{\infty}\frac{(-1)^n}{n-1}\sum_{k=0}^{n}(3n-k)^j{n\choose k}A^{n-k}B^k\;, \end{equation} being $j\gg1$ an integer number and $0\leq A,B\leq 1$. In order to deal with this expression, my idea was to approximate the inner sum and to convert the outer one into an integral (evaluating it maybe numerically if there are no further approximations). Thanks for any help.

Answer 1

Note that the outer sum cannot start from $n=1$, as the denominator is zero there. This said, you can write $$ (3n-k)^j=\frac{\partial^j}{\partial s^j}e^{s(3n-k)}\Big|_{s\to 0}\ . $$ Therefore \begin{equation} \sum_{k=0}^{n}(3n-k)^j{n\choose k}A^{n-k}B^k=\frac{\partial^j}{\partial s^j} \sum_{k=0}^{n}e^{s(3n-k)}{n\choose k}A^{n-k}B^k\Big|_{s\to 0}=\frac{\partial^j}{\partial s^j}\left[A^n e^{3 n s} \left(\frac{B e^{-s}}{A}+1\right)^n\right]\Big|_{s\to 0}\ , \end{equation} and (assuming your outer sum starts from $n=2$) one has \begin{equation} \frac{\partial^j}{\partial s^j}\sum_{n=2}^{\infty}\frac{(-1)^n}{n-1}\left[A^n e^{3 n s} \left(\frac{B e^{-s}}{A}+1\right)^n\right]\Big|_{s\to 0}=\frac{\partial^j}{\partial s^j}\left[e^{2 s} \left(A e^s+B\right) \log \left(A e^{3 s}+B e^{2 s}+1\right)\right]\Big|_{s\to 0}\ , \end{equation} where one uses $\sum_{n=2}^\infty \frac{(-1)^n z^n}{n-1}=z \log(1+z)$. I checked that (for the outer sum starting from $n=2$), setting $j=5$, $A=1/3$ and $B=1/6$, your initial sum is $\approx 371.369...$ and the exact result (taking five derivatives) is $\frac{6620645}{19683}+\frac{259}{3} \log \left(\frac{3}{2}\right)\approx 371.369...$