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:

$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$$

-
your sum is finite... –  draks ... Jul 7 '14 at 8:47
@draks... yes... I curious to know that... –  zardav Jul 7 '14 at 8:50
I like how the top two answers, at least at the moment, are virtually identical... –  Bair Jul 7 '14 at 21:43

Hint:
$$\frac{1}{(k+1)(k+2)} = \frac{1}{k+1}-\frac{1}{k+2}$$

-

Hint: $$\frac 1 {(k+1)(k+2)}=\frac {1} {k+1}-\frac {1} {k+2}$$

-
@Elimination beat you to it by 22 seconds. –  Dan Jul 8 '14 at 4:20
It says he beat you to it? –  Lost1 Jul 20 '14 at 19:22

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!

-
typo: in the very first expression the argument is $k$, not $n$ –  Alex Jul 7 '14 at 9:03
Thanks. Fixed.. –  zardav Jul 7 '14 at 9:05

Hints: 1) expand the partial fractions, 2) use the telescoping sum 3) take the limit

-

We can use integrals to calculate this sum: $$\sum_{k=1}^{n}\dfrac{1}{(k+1)(k+2)} = \sum_{k=1}^{n}\biggl(\dfrac{1}{k+1} - \dfrac{1}{k+2}\biggr) = \sum_{k=1}^{n}\biggl(\int_{0}^{1}x^kdx - \int_{0}^{1}x^{k+1}dx \biggr)$$ $$=\sum_{k=1}^{n}\int_{0}^{1}x^k(1 - x)dx = \int_{0}^{1}(1 - x)\sum_{k=1}^{n}x^kdx = \int_{0}^{1}(1 - x)\dfrac{1 - x^{n+1}}{1 - x}dx$$ $$= \int_{0}^{1}(1 - x^{n+1})dx = \biggl[\dfrac{x^2}{2} - \dfrac{x^{n+2}}{n+2}\biggr]_{0}^{1} = \dfrac{1}{2} - \dfrac{1}{n + 2}$$

-

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

-