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:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$$
 A: Hints: 1) expand the partial fractions, 2) use the telescoping sum 3) take the limit 
A: 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:
$$\begin{array}{l}\sum\limits_{k = 1}^\infty  {\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}}  = \frac{1}{{2 - 1}}\frac{1}{{1 + 1}} = \frac{1}{2}\end{array}
$$
A: Hint:
$$\frac{1}{(k+1)(k+2)} = \frac{1}{k+1}-\frac{1}{k+2}$$
A: Hint:  $$\frac 1 {(k+1)(k+2)}=\frac {1} {k+1}-\frac {1} {k+2}$$
A: OK, I answer the question with the hint:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)} = \sum_{k=1}^n \left(\frac 1 {k+1} - \frac 1 {k+2}\right) = \\ = \left( \frac 1 2 - \frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) + \left( \frac 1 4 - \frac 1 5 \right) + \ldots + \left( \frac 1 {n+1} - \frac 1 {n+2} \right) = \\ = \frac 1 2 - \frac 1 {n+2}$$
(For my homework: $\lim_{n\to\infty} \frac 1 2 - \frac 1 {n+2} = \frac 1 2$)
Thanks!
A: We can use integrals to calculate this sum:
$$
\sum_{k=1}^{n}\dfrac{1}{(k+1)(k+2)} = \sum_{k=1}^{n}\biggl(\dfrac{1}{k+1} - \dfrac{1}{k+2}\biggr) = \sum_{k=1}^{n}\biggl(\int_{0}^{1}x^kdx - \int_{0}^{1}x^{k+1}dx  \biggr)
$$
$$
=\sum_{k=1}^{n}\int_{0}^{1}x^k(1 - x)dx =  \int_{0}^{1}(1 - x)\sum_{k=1}^{n}x^kdx = \int_{0}^{1}(1 - x)\dfrac{x - x^{n+1}}{1 - x}dx
$$
$$
= \int_{0}^{1}(x - x^{n+1})dx = \biggl[\dfrac{x^2}{2} - \dfrac{x^{n+2}}{n+2}\biggr]_{0}^{1} = \dfrac{1}{2} - \dfrac{1}{n + 2}
$$
