Calculate this infinite sum $$s= \sum_{n=1}^\infty \frac{n+3}{(2^n)(n+1)(n+2)}$$
Any method to calculate this type of infinite sums?

Using Partial Fraction Decomposition,
$$\text{let }\frac{n+3}{(n+1)(n+2)}=\frac A{n+1}+\frac B{n+2}$$
$$n+3=n(A+B)+2A+B\implies A+B=1,2A+B=3\implies A=2, B=-1$$
$$\implies\frac{n+3}{(n+1)(n+2)}=\frac 2{n+1}-\frac1{n+2}$$
$$\implies\frac{n+3}{2^n(n+1)(n+2)}=\frac{(1/2)^{n-1}}{n+1}-\frac{(1/2)^n}{n+2}$$
If $T_m=\dfrac{(1/2)^{m-1}}{m+1},$
$$\frac{n+3}{2^n(n+1)(n+2)}=T(n)-T(n+1)$$ which is clearly Telescoping 
