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.
WLOG
$$\dfrac r{(r+1)(r+3)(r+4)}=a\left(\dfrac1{r+1}-\dfrac1{r+3}\right)+b\left(\dfrac1{r+3}-\dfrac1{r+4}\right)\  \ \ \  (1)$$
so that
$$f(n)=\sum_{r=1}^n\dfrac r{(r+1)(r+3)(r+4)}$$
$$=a\sum_{r=1}^n\left(\dfrac1{r+1}-\dfrac1{r+3}\right)+b\sum_{r=1}^n\left(\dfrac1{r+3}-\dfrac1{r+4}\right)$$
$$=a\left(\dfrac12-\dfrac14+\dfrac13-\dfrac15+\cdots+?\right)+b\left(\dfrac14-\dfrac1{n+4}\right)$$
$$\lim_{n\to\infty}f(n)=a\left(\dfrac12-\dfrac14+\dfrac13-\dfrac15\right)+b\left(\dfrac14\right)$$
Now from $(1),$  $$r=2a(r+4)+b(r+1)=r(2a+b)+8a+b$$
Now compare the coefficients of $r$ and constant to find $$8a+b=0\ \ \ \ (2), 2a+b=1\ \ \ \ (3)$$
Finally solve the two simultaneous solutions to find $a,b$