Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

share|improve this question
3  
your sum is finite... –  draks ... Jul 7 at 8:47
    
@draks... yes... I curious to know that... –  zardav Jul 7 at 8:50
6  
I like how the top two answers, at least at the moment, are virtually identical... –  Bair Jul 7 at 21:43

6 Answers 6

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

share|improve this answer

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

share|improve this answer
    
@Elimination beat you to it by 22 seconds. –  Dan Jul 8 at 4:20
    
It says he beat you to it? –  Lost1 Jul 20 at 19:22
up vote 11 down vote accepted

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!

share|improve this answer
    
typo: in the very first expression the argument is $k$, not $n$ –  Alex Jul 7 at 9:03
    
Thanks. Fixed.. –  zardav Jul 7 at 9:05

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

share|improve this answer

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

share|improve this answer

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: $$\begin{array}{l}\sum\limits_{k = 1}^\infty {\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}} = \frac{1}{{2 - 1}}\frac{1}{{1 + 1}} = \frac{1}{2}\end{array} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.