Sum outta nowhere!

We know $\Gamma(k) = (k-1)!$ when $k$ is a positive integer.

So we have the summation as :-

$\large -\ln(7).\sum_{k=0}^{\infty} (-1)^{k}\frac{(\ln(7))^{k}(k+1)^{2}}{k!} ..........................................(1)$

We have the Mcalurin expansion of $e^{x}$ as

$\large e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!}$

So multiplying by $x$ on both sides and then differentiating once wrt $x$ on both sides we have

$\large xe^{x} + e^{x} = \sum_{k=0}^{\infty} \frac{(k+1)x^{k}}{k!}$

again doing the same thing we have :-

$\large x^{2}e^{x} + 3xe^{x} + e^{x} = \sum_{k=0}^{\infty} \frac{(k+1)^{2}x^{k}}{k!}..............................................(2)$

So to connect $(1)$ and $(2)$

We multiply $(2)$ by $-\ln(7)$ and in place of $x$ we put $x=-\ln(7)$

So we have our result as :-

$\large \frac{1}{7}(-(\ln(7))^{3} +3(\ln(7))^{2} - \ln(7))$

So arranging $A,B,C,D,E,F$ as required by the problem we have the number

$ABCDEF = 171323$