Back to summation

$\frac { 1 }{ 3 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n(n+1)(n+2)(n+3) } } \times \left\{ (n+3)-n \right\}$

$=\frac { 1 }{ 3 } \left[ \sum _{ n=1 }^{ \infty }{ \left\{ \frac { 1 }{ n(n+1)(n+2) } -\frac { 1 }{ (n+1)(n+2)(n+3) } \right\} } \right]$

All the intermediate terms gets cancelled

$=\frac { 1 }{ 3 } \left[ \frac { 1 }{ 1.2.3 } \right] =\frac { 1 }{ 18 }$ .

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{n(n+1)(n+2)(n+3)} & \small \color{#3D99F6} \text{Using partial fractions} \\ & = \sum_{n=1}^\infty \left(\frac 1{6n} - \frac 1{2(n+1)} + \frac 1{2(n+2)} - \frac 1{6(n+3)} \right) \\ & = \frac 16 \sum_{n=1}^\infty \left(\frac 1{n} - \frac 3{n+1} + \frac 3{n+2} - \frac 1{n+3} \right) \\ & = \frac 16 \left(\frac 11 + \frac 12 + \frac 13 - \frac 32 \right) \\ & = \frac 1{18} \end{aligned}$

$\implies A+B = 1+18 = \boxed{19}$

$\displaystyle \sum_{n=1}^{\infty} \dfrac 1 {n(n+1)(n+2)(n+3)}=\sum_{n=1}^{\infty} \dfrac {(n+3)-n}{n(n+1)(n+2)(n+3)}\\ \displaystyle {\color{#D61F06}{ \sum_{n=1}^{\infty} \dfrac 1 {3*n(n+1)(n+2)} } }-\sum_{n=1}^{\infty} \dfrac 1 {3*(n+1)(n+2)(n+3)}\\ \text{Only term with n=1 in the red series remain. All other terms cancels out.}\\ \therefore~\displaystyle \frac 1 3* \sum_{n=1}^1 \dfrac 1 {n(n+1)(n+2)} = \dfrac 1 {3*(1*2*3)}=\frac A B.~~~~~~So~~A+B=1+18=19.$