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.
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{N \in \mathbb{N}_{\ \geq\ 1}}$:
\begin{align}
&\bbox[10px,#ffd]{\sum_{r = 1}^{N}{r \over
\pars{r + 1}\pars{r + 3}\pars{r + 4}}}
\\[5mm] = &\
-\,{1 \over 6}\sum_{r = 1}^{N}{1 \over r + 1} +
{1 \over 6}\sum_{r = 1}^{N}{1 \over r + 3} -
{4 \over 3}\sum_{r = 1}^{N}{1 \over r + 4}
\\[5mm] = &\
-\,{1 \over 6}\sum_{r = 2}^{N + 1}{1 \over r} +
{3 \over 2}\sum_{r = 4}^{N + 3}{1 \over r} -
{4 \over 3}\sum_{r = 5}^{N + 4}{1 \over r}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
-\,{1 \over 6}\pars{-1} + {3 \over 2}\pars{-1 - {1 \over 2} - {1 \over 3}}
-{4 \over 3}\pars{-1 - {1 \over 2} - {1 \over 3} - {1 \over 4}}
\end{align}
Note that $\ds{-1/6 + 3/2 - 4/3 = \color{red}{\large 0}}$.
\begin{align}
&\bbox[10px,#ffd]{\sum_{r = 1}^{\infty}{r \over
\pars{r + 1}\pars{r + 3}\pars{r + 4}}} =
{1 \over 6} - {11 \over 4} + {25 \over 9} = \bbx{7 \over 36}
\approx 0.1944
\\ &\ \mbox{}
\end{align}
A: Hint: Use this here $$\frac{3}{2 (r+3)}-\frac{4}{3 (r+4)}-\frac{1}{6 (r+1)}$$
A: 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.
A: 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 $a,b$
