Sum this Factorial Series

$\begin{aligned} S & = 1^2\cdot \frac {1!}{5!} + 2^2\cdot \frac {2!}{6!} + 3^2\cdot \frac {3!}{7!} + \cdots \\ & = \sum_{k=1}^\infty \frac {k^2k!}{(k+4)!} \\ & = \sum_{k=1}^\infty \frac {k^2}{(k+1)(k+2)(k+3)(k+4)} & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \frac 16 \sum_{k=1}^\infty \left(\frac 1{k+1} - \frac {12}{k+2} + \frac {27}{k+3} - \frac {16}{k+4}\right) \\ & = \frac 16 \left(\sum_{k=2}^\infty \frac 1k - 12 \sum_{k=3}^\infty \frac 1k + 27\sum_{k=4}^\infty \frac 1k - 16 \sum_{k=5}^\infty \frac 1k \right) \\ & = \frac 16 \left(\frac 12 + \frac 13 + \frac 14 + \sum_{k=5}^\infty \frac 1k - \frac {12}3 - \frac {12}4 - 12 \sum_{k=5}^\infty \frac 1k + \frac {27}4 + 27\sum_{k=5}^\infty \frac 1k - 16 \sum_{k=5}^\infty \frac 1k \right) \\ & = \frac 16 \left(\frac 12 + \frac 13 + \frac 14 - \frac {12}3 - \frac {12}4 + \frac {27}4 \right) \\ & = \frac 5{36} \end{aligned}$

Therefore, $a+b = 5 + 36 = \boxed{41}$ .

$S=\sum _{ k=1 }^{ \infty }{ \frac { k^{ 2 }\cdot k! }{ \left( k+4 \right) ! } } =\sum _{ k=1 }^{ \infty }{ \frac { k^{ 2 } }{ \left( k+1 \right) \left( k+2 \right) \left( k+3 \right) \left( k+4 \right) } } \\ \\ S=\frac { 1 }{ 6 } \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ \left( k+1 \right) } -\frac { 12 }{ \left( k+2 \right) } +\frac { 27 }{ \left( k+3 \right) } -\frac { 16 }{ \left( k+4 \right) } } \\ \\ S=\frac { 1 }{ 6 } \sum _{ k=1 }^{ \infty }{ \int _{ 0 }^{ 1 }{ \left( x^{ k }-12x^{ k+1 }+27x^{ k+2 }-16x^{ k+3 } \right) dx } } \\ \\ S=\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ \sum _{ k=1 }^{ \infty }{ \left( x^{ k }-12x^{ k+1 }+27x^{ k+2 }-16x^{ k+3 } \right) } dx } \\ \\ S=\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ \left( 1-12x+27x^{ 2 }-16x^{ 3 } \right) \sum _{ k=1 }^{ \infty }{ x^{ k } } dx } \\ \\ S=\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ \left( 16{ x }^{ 2 }-11x+1 \right) \left( 1-x \right) \frac { x }{ 1-x } dx } =\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ \left( 16{ x }^{ 3 }-11x^{ 2 }+x \right) dx } \\ \\ S=\frac { 1 }{ 6 } \left( 4-\frac { 11 }{ 3 } +\frac { 1 }{ 2 } \right) =\frac { 5 }{ 36 }$

$\begin{aligned} & S_n = 1^2 . \frac{1!}{5!} + 2^2. \frac{2!}{6!} + 3^2.\frac{3!}{7!} +4^2 . \frac{4!}{8!} +\cdots \\& S_n = \displaystyle\sum_{n=1}^{\infty} \dfrac{n^2.n!}{(n+4)!} =\displaystyle\sum_{n=1}^{\infty}\dfrac{n^2.n!}{(n+4)(n+3)(n+2)(n+1)n!} \\& S_n = \displaystyle\sum_{n=0}^{\infty}\dfrac{n^2}{(n+4)(n+3)(n+2)(n+1)} \\ & S_n = \displaystyle\sum_{n=1}^{\infty} \dfrac{(n^2 -2^2 )+2^2}{(n+4)(n+3)(n+2)(n+1)} \\& S_n = \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{n-2}{(n+4)(n+3)(n+1)} +\dfrac{4}{(n+1)(n+2)(n+3)(n+4)} \right) \\& S _n = \displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{(n+4)(n+3)} -\dfrac{3}{(n+4)(n+3)(n+1)} +\dfrac{2}{(n+1)(n+4)} - \dfrac{2}{(n+3)(n+2)}\right) \\& S_n = \displaystyle\sum_{n=1}^{\infty} \left( \dfrac{1}{n+3} -\dfrac{1}{n+4} - \dfrac{1}{n+3}\left(\dfrac{1}{n+1} -\dfrac{1}{n+4}\right) +\dfrac{2}{3}\left(\dfrac{1}{n+1} - \dfrac{1}{n+4} \right)-2 \left(\dfrac{1}{n+2} -\dfrac{1}{n+3}\right)\right)\\& S_n =\displaystyle\sum_{n=0}^{\infty}\left(\dfrac{1}{n+3}-\dfrac{1}{n+4} - \dfrac{1}{(n+3)(n+1)} +\dfrac{1}{(n+3)(n+4)}+\dfrac{2}{3(n+1)} -\dfrac{2}{3(n+4)}-\dfrac{2}{(n+2)}+\dfrac{2}{n+3}\right)\\& S_n = \displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{n+3} -\dfrac{1}{n+4} -\dfrac{1}{2(n+1)} +\dfrac{1}{2(n+3)}+ \dfrac{1}{n+3} -\dfrac{1}{n+4}+\dfrac{2}{3(n+1)} -\dfrac{2}{3(n+4)}-\dfrac{2}{(n+2)}+\dfrac{2}{n+3}\right)\\& S_n = \displaystyle\sum_{n=1}^{\infty} \left[\left({\color{#3D99F6}\dfrac{1}{n+3}+\dfrac{1}{2(n+3)} +\dfrac{1}{n+3}+ \dfrac{2}{n+3}}\right)+\left({\color{#D61F06}-\dfrac{1}{n+4} -\dfrac{1}{n+4} - \dfrac{2}{3(n+4)}}\right)+\left(\dfrac{2}{3(n+1)} - \dfrac{1}{2(n+1)}\right) - \dfrac{2}{n+2} \right]\\& S_n = \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{9}{2(n+3)} -\dfrac{8}{3(n+4)} + \dfrac{1}{6(n+1)} -\dfrac{2}{n+2} \right) \\& S_n = \dfrac{1}{6}\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)} -\dfrac{12}{n+2} + \dfrac{27}{(n+3)} - \dfrac{16}{(n+4)}\right) \\& S_n = \dfrac{1}{6}\displaystyle\sum_{n=1}^{\infty}\left[12\left(\frac{1}{n+1} - \frac{1}{n+2}\right) +16\left(\dfrac{1}{n+3} -\dfrac{1}{n+4} \right) +\left(-\dfrac{11}{n+1} +\dfrac{11}{n+3} \right)\right] \\& S_n = \dfrac{1}{6}\left(6 + 4 -11(\frac{1}{2} + \frac{1}{3})\right) \\& S_n = \dfrac{1}{6}\times \dfrac{60-55}{6}=\boxed{ \dfrac{5}{36} }\end{aligned}$