Method of difference (Series) 
Find $$\sum_{r=1}^{n}\frac{1}{r(r+3)}$$ by the method of differencing.

My attempt, 
$$\sum_{r=1}^{n}\frac{1}{r(r+3)}=\sum_{r=1}^{n}\left(\frac{1}{3r}-\frac{1}{3(r+3)}\right)$$
$$=\left(\frac{1}{3}-\frac{1}{12}\right)+\left(\frac{1}6-\frac{1}{15}\right)+\dots$$
This question is different as I cant see the common fraction. Hope someone can point it out.
 A: $$\require{cancel}
\begin{align}
\sum_{r=1}^n\frac 1{r(r+3)}
&=\frac 13\sum_{r=1}^n\frac 1r-\frac 1{r+3}\\
&=\frac 13 \bigg[\bigg(\color{red}{\frac 11}-\cancel{\frac 14}\bigg)\\
&\;\;\quad+\bigg(\color{red}{\frac 12}-\cancel{\frac 15}\bigg)\\
&\;\;\quad+\bigg(\color{red}{\frac 13}-\cancel{\frac 16}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 14}-\cancel{\frac 17}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 15}-\cancel{\frac 18}\bigg)\\
&\qquad\qquad\vdots\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-3}}-\cancel{\frac 1n}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-2}}\color{blue}{-\frac 1{n+1}}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-1}}\color{blue}{-\frac 1{n+2}}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n}}\color{blue}{-\frac 1{n+3}}\bigg)\bigg]\\
&=\frac 13\bigg[\color{red}{1+\frac 12+\frac 13}\color{blue}{-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}}\bigg)\bigg]\\
&=\frac 13\bigg[\frac {11}6-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}\bigg)\bigg]
\end{align}$$

Alternatively, in less graphical fashion, 
$$\begin{align}
\sum_{r=1}^n\frac 1{r(r+3)}
&=\frac 13\bigg[\sum_{r=1}^n\frac 1r-\sum_{r=1}^n\frac 1{r+3}\bigg]\\
&=\frac 13\bigg[\sum_{r=1}^n \frac 1r-\sum_{r=4}^{n+3}\frac 1r\bigg]\\
&=\frac 13\bigg[\bigg(\sum_{r=1}^3 \frac 1r+\sum_{r=4}^n \frac 1r\bigg)-\bigg(\sum_{r=4}^n \frac 1r+\sum_{r=n+1}^{n+3} \frac 1r\bigg)\bigg]\\
&=\frac 13\bigg[\sum_{r=1}^3 \frac 1r-\sum_{r=n+1}^{n+3}\frac 1r\bigg]\\
&=\frac 13\bigg[\frac {11}6-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}\bigg)\bigg]
\end{align}$$
A: One does not need to work with a general method. It might be helpful to observe it in a different way. The factor $\frac13$ is really irrelevant. Now,
$$
\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\frac13+\left(\frac14+\frac15+\frac16+\cdots+\frac{1}{n}\right)
$$
$$
\sum_{k=1}^n\frac{1}{k+3}= \left(\frac14+\frac15+\frac16+\cdots+\frac{1}{n}\right)+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}
$$
Both are finite sums. Just do the subtraction in a very obvious way to get
$$
\sum_{k=1}^n\frac{1}{k(k+3)}=\frac13\left(\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\frac{1}{k+3}\right)
$$
A: Hint. One may write
$$
\frac{1}{r(r+3)}=\frac13\left(\frac{1}{r}-\frac{1}{(r+1)}\right)\color{red}{+}\frac13\left(\frac{1}{(r+1)}-\frac{1}{(r+2)}\right)\color{red}{+}\frac13\left(\frac{1}{(r+2)}-\frac{1}{(r+3)}\right)
$$ then one may use telescoping terms.
A: Try writing out a few more terms (at least the first 5 and last 5 pairs). You should find that most of the terms cancel and you're left with only a finite number of terms.
