Calculate $\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$ I have homework questions to calculate infinity sum, and when I write it into wolfram, it knows to calculate partial sum... 
So... How can I calculate this:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$$

Lets solve the problem generally;
$$\begin{array}{l}\sum\limits_{k = i}^\infty  {\frac{1}{{\left( {k + a} \right)\left( {k + b} \right)}}}  = \\\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = i}^n {\frac{1}{{\left( {k + a} \right)\left( {k + b} \right)}}}  = \\\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = i}^n {\frac{1}{{b - a}}\left[ {\frac{1}{{k + a}} - \frac{1}{{k + b}}} \right]}  = \\\frac{1}{{b - a}}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = i}^n {\left[ {\frac{1}{{k + a}} - \frac{1}{{k + b}}} \right]} = \\\frac{1}{{b - a}}\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{i + a}} - \frac{1}{{n + b}}} \right]\end{array}
$$
So the solution is:
$$\begin{array}{l}\sum\limits_{k = i}^\infty  {\frac{1}{{\left( {k + a} \right)\left( {k + b} \right)}}}  = \frac{1}{{b - a}}\frac{1}{{i + a}}\end{array}
$$
And going back to your questions:
$$\begin{array}{l}\sum\limits_{k = 1}^\infty  {\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}}  = \frac{1}{{2 - 1}}\frac{1}{{1 + 1}} = \frac{1}{2}\end{array}
$$
