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.

The partial fraction expansion of your summand is
$$-\frac{1}{6}\frac{1}{r+1} + \frac{3}{2}\frac{1}{r+3} - \frac{4}{3}\frac{1}{r+4}.$$
Then notice that $3/2 = 4/3+1/6$ so you have
$$-\frac{1}{6}\frac{1}{r+1} + \frac{1}{6}\frac{1}{r+3}+\frac{4}{3}\frac{1}{r+3} - \frac{4}{3}\frac{1}{r+4}.$$
And now things telescope like crazy.
