Find the sum of the series with terms given by ${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$ The given series has general term as $${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$$
I have tried to approach this problem by making a telescopic series as follows, but I end up cancelling $r$ in the numerator, $$\frac{1}{(r+1)(r+3)}-\frac{1}{(r+3)(r+4)}=\frac{3}{(r+1)(r+3)(r+4)}$$ Please provide an alternate approach to form telescopic series.

In general you can guess (in general more hope) a decomposition $\frac{r}{(r+1)(r+3)(3+4)} = \frac{a}{r+1}+\frac{b}{r+3} + \frac{c}{r+4}$ for certain $a,b,c \in \mathbb{R}$.
So what you have to do is to solve the system given by $ \frac{r}{(r+1)(r+3)(3+4)} = \frac{a}{r+1}+\frac{b}{r+3} + \frac{c}{r+4} = \frac{a(r+3)(r+4) + b(r+1)(r+4) + c(r+1)(r+3)}{(r+1)(r+3)(r+4)}$.
Basically you will have to impose that the leading coefficient of $r^{2}$ will be $0$, as well as the costant term, and the one of $r$ equals to $1$.
This traslates into a linear system. If my calculus are correct we get $a(r^{2}+7r+12) + b(r^{2}+5r+4)+ c(r^{2}+4r+3) = r^{2}(a+b+c) + r(7a+5b+4c)+12a+4b+3c$. The system now becomes $\begin{cases}a+b+c = 0 \\ 7a+5b+4c = 1 \\ 12a+4b+3c = 0\end{cases}$. I'm more comfort evaluating the system as $\begin{pmatrix} 1 & 1 & 1 \ 7 & 5 & 4 \ 12 & 4 & 3\end{pmatrix}\begin{pmatrix} a\ b \ c\end{pmatrix} = \begin{pmatrix} 0\ 1 \ 0\end{pmatrix}
Since the matrix is invertible (the determinant should be nonzero if I'm not mistaken) it surely exists $a,b,c \in \mathbb{R}$ as required, so you can proceed in solving the system to determine them which will link to other answers given.
