Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$ $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.
Do I have to use something else? Or am I missing some easy manipulation?

There still is the brutal way:
$$ \frac{r+2}{r(r+1)(r+3)}\stackrel{\text{PFD}}{=}\frac{2}{3}\cdot\frac{1}{r}-\frac{1}{2}\cdot\frac{1}{r+1}-\frac{1}{6}\cdot\frac{1}{r+3} \tag{1}$$
leads to:
$$ \frac{r+2}{r(r+1)(r+3)} = \frac{1}{6}\int_{0}^{1} x^{r-1}\left(4-3x-x^3\right)\,dx \tag{2}$$
then by summing both sides over $r\geq 1$:
$$ \sum_{r\geq 1}\frac{r+2}{r(r+1)(r+3)}=\frac{1}{6}\int_{0}^{1}\frac{4-3x-x^3}{1-x}\,dx=\frac{1}{6}\int_{0}^{1}\left(4+x+x^2\right)\,dx \tag{3} $$
and simplifying:
$$ \sum_{r\geq 1}\frac{r+2}{r(r+1)(r+3)}=\frac{1}{6}\left(4+\frac{1}{2}+\frac{1}{3}\right)=\color{red}{\frac{29}{36}}.\tag{4}$$
The partial sums are more directly computed via $(1)$:
$$\begin{eqnarray*} \sum_{r=1}^{n}\frac{r+2}{r(r+1)(r+3)}&=&\frac{2}{3}H_n-\frac{1}{2}\left(H_{n+1}-1\right)-\frac{1}{6}\left(H_{n+3}-\frac{11}{6}\right)\\&=&\frac{157 n+138 n^2+29 n^3}{36 (n+1)(n+2)(n+3)}.\tag{5} \end{eqnarray*}$$
