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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:

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

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    $\begingroup$ your sum is finite... $\endgroup$ – draks ... Jul 7 '14 at 8:47
  • $\begingroup$ @draks... yes... I curious to know that... $\endgroup$ – zardav Jul 7 '14 at 8:50
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    $\begingroup$ I like how the top two answers, at least at the moment, are virtually identical... $\endgroup$ – Dair Jul 7 '14 at 21:43
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Hint:
$$\frac{1}{(k+1)(k+2)} = \frac{1}{k+1}-\frac{1}{k+2}$$

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

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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!

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

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  • $\begingroup$ In the last line you took $x^2/2$ as antiderivative of the constant $1$ function. Interesting that nobody noticed that mistake so far. $\endgroup$ – principal-ideal-domain Aug 26 '15 at 14:21
  • $\begingroup$ Another mistake is the second line where you write $\sum_{k=1}^n x^k$ as quotient. Note that the sum starts at $k=1$ and not $k=0$. $\endgroup$ – principal-ideal-domain Aug 26 '15 at 14:28
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Hints: 1) expand the partial fractions, 2) use the telescoping sum 3) take the limit

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

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