Almost Like A Factorial Denominator

$\large S=\displaystyle \sum_{n=1}^{\infty} \dfrac{n+3}{n(n+1)(n+2)}$ $=\displaystyle \sum_{n=1}^{\infty} \dfrac{\not n}{\not n(n+1)(n+2)}+\displaystyle \sum_{n=1}^{\infty} \dfrac{3}{n(n+1)(n+2)}$ $(\displaystyle \sum_{n=1}^{\infty} (\dfrac{1}{n+1}- \dfrac{1}{n+2}))~~+(\dfrac{3}{2}~(\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}))\\~~~~\color{goldenrod}{(Both ~are~Telescopic~Series)}$ $=\large \dfrac{1}{2}+\dfrac{3}{4}=\boxed{\color{#D61F06}{\boxed {1.25}}}$

$\displaystyle f(n)=\sum_1^{\infty}\dfrac {n+3}{n*(n+1)*(n+2)}=\sum_1^{\infty}\dfrac {2(n+2) - (n+1)}{(n*(n+1)*(n+2)}\\ \displaystyle f(n)=\sum_1^{\infty}\dfrac 2 {n*(n+1)} - \sum_1^{\infty}\dfrac 1{(n*(n+2)}.\\ \dfrac 2 {n*(n+1)} =\dfrac A n + \dfrac B {n+1} ~~~~and~~~ \dfrac 1{n*(n+2)}=\dfrac C n + \dfrac D {n+2} .\\ \text{Solving for A, B, C, D, and combining A and C}\\ \displaystyle f(n)=\dfrac 3 2 *\sum_1^{\infty} \dfrac 1 n +\dfrac1 2 *\sum_1^{\infty} \dfrac 1{ n+2} - 2 *\sum_1^{\infty} \dfrac 1{n +1}\\ \small \text{We could obtain partial fractions directly by solving 3 equations in three unknowns.}\\ \small \text{ But here we have split into two sets of 2 equations in 2 unknowns, I felt that is easy.} \\ Also ~ refresh:- \displaystyle \sum_{n=a}^b f(n)=\sum_{n=a - c}^{b - c} f(n + c) , ~~our ~b=\infty.\\ \therefore \displaystyle f(n)=\dfrac 3 2 *\sum_{n=1}^{\infty} \dfrac 1 n + \dfrac1 2 *\sum_{n=3}^{\infty} \dfrac 1 n - 2 *\sum_{n=2}^{\infty} \dfrac 1n \\ \displaystyle \therefore \color{#3D99F6}{\dfrac 3 2 *\sum_{n=1}^2 \dfrac 1 n} + \dfrac 3 2 *\sum_{n=3}^{\infty} \dfrac 1 n+ \dfrac1 2 *\sum_{n=3}^{\infty} \dfrac 1 n - \color{#3D99F6}{ 2 *\sum_{n=2}^2 \dfrac 1n} -2 *\sum_{n=3}^{\infty} \dfrac 1n \\ \text{In telescoping series f(n) the only terms those does not cancel out are }\\ \displaystyle~ \color{#3D99F6}{\dfrac 3 2 *\sum_{n=1}^2 \dfrac 1 n} - \color{#3D99F6}{2 *\sum_{n=2}^2\dfrac 1 n} = \dfrac 3 2 *( \dfrac 1 1 + \dfrac 1 2)- 2*\dfrac 1 2= \dfrac 5 4=\Large ~~\color{#D61F06}{1.25}$