Series exercise asked by Mohamed Mathe in the Lebanese university math group

Exercise:

Find $\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+2k}}$

Solution: we have ${{k}^{2}}+2k=k\left( k+2 \right)$

So $\frac{1}{{{k}^{2}}+2k}=\frac{1}{k\left( k+2 \right)}=\frac{A}{k}+\frac{B}{k+2}=\frac{\left( A+B \right)k+2A}{k\left( k+2 \right)}$

Thus $A+B=0\,\,\,,\,\,2A=1\Leftrightarrow A=\frac{1}{2}\,\,\And \,\,B=-\frac{1}{2}$

Hence $\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+2k}=\sum\limits_{k=1}^{n}{\frac{1}{k\left( k+2 \right)}=\frac{1}{2}\left( \sum\limits_{k=1}^{n}{\frac{1}{k}-\sum\limits_{k=1}^{n}{\frac{1}{k+2}}} \right)}}$

$=\frac{1}{2}\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{n-1}+\frac{1}{n}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-...-\frac{1}{n-1}-\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2} \right)$

$=\frac{1}{2}\left( 1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2} \right)=\frac{1}{2}\left( \frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2} \right)=\frac{n\left( 3n+5 \right)}{4\left( n+1 \right)\left( n+2 \right)}$