TITLE: Finding $b-a$ for positive integer $a$, $b$ satisfying $\sum_{x=1}^\infty\frac{3x^2+12x+16}{(x(x+1)(x+2)(x+3)( x+4))^3}=\frac1{4(a!)^b}$
QUESTION [6 upvotes]: $$
\sum_{x = 1}^{\infty}\frac{3x^{2} + 12x + 16}
{\left[\vphantom{A^{A}}x\left(x + 1\right)\left(x + 2\right)
\left(x + 3\right)\left(x + 4\right)\right]^{\, 3}} =
\frac{1}{4\left(a!\right)^{b}}
$$
Compute $b-a$ if $b$ and $a$ are positive integers.

I asked my teacher to help me in solving this sum. But, unfortunately, he said I can't. it is a very hard question, So, I hope you can help me in solving this problem.

REPLY [5 votes]: The answer is $\color{green}{-1}$.
Indeed, the sum converges very quickly (the terms are $\Theta(x^{-13})$) and the inverses of the partial sums are
$$55741.9354839\\
55312.3888560\\
55297.2897327\\
55296.1612133\\
55296.0276428\\
55296.0059708\\
55296.0015387\\
55296.0004559\\
55296.0001513\\
55296.0000551\\
\cdots$$
Obviously, this tends to $4\cdot4!^3$.