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I'm looking to calculate the sum of the following infinite series:

$$\sum^\infty_{n=0} \frac{1}{(n+1)(n+5)}$$

and I'm not too sure where to begin. There doesn't seem to be a common ratio. I have tried working it out by dividing it into partial fractions, but this got me nowhere. What should I do?

Thanks.

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    $\begingroup$ telescoping sum? use partial fractions. $\endgroup$
    – abel
    Mar 1, 2015 at 1:47
  • $\begingroup$ How should I go about doing that? $\endgroup$ Mar 1, 2015 at 1:48

2 Answers 2

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$\frac1{(n+1)(n+5)} =\frac14 (\frac1{n+1}-\frac1{n+5}) $.

Use this to get $\sum_{n=1}^m \frac1{(n+1)(n+5)} $ and let $m \to \infty$.

(added later to explain more fully)

Let $s_m =\sum_{n=1}^m \frac1{(n+1)(n+5)} $, where $m > 5$. Then

$\begin{array}\\ s_m &=\sum_{n=1}^m \frac1{(n+1)(n+5)}\\ &=\frac14 \sum_{n=1}^m (\frac1{n+1}-\frac1{n+5})\\ &=\frac14 \sum_{n=1}^m \frac1{n+1}- \frac14 \sum_{n=1}^m \frac1{n+5}\\ &=\frac14 \sum_{n=2}^{m+1} \frac1{n}- \frac14 \sum_{n=6}^{m+5} \frac1{n}\\ &=\frac14 \left(\sum_{n=2}^{m+1} \frac1{n}- \sum_{n=6}^{m+5} \frac1{n}\right)\\ &=\frac14 \left(\sum_{n=2}^{5} \frac1{n}+\sum_{n=6}^{m+1} \frac1{n}\right)- \frac14 \left(\sum_{n=6}^{m+1} \frac1{n}+\sum_{n=m+2}^{m+5} \frac1{n}\right)\\ &=\frac14 \left(\sum_{n=2}^{5} \frac1{n}\right)- \frac14 \left(\sum_{n=m+2}^{m+5} \frac1{n}\right) \quad\text{(the common sums cancel)}\\ \end{array} $

Since $\lim_{m \to \infty} \sum_{n=m+2}^{m+5} \frac1{n} = 0 $ (since it is less than $\frac{4}{m}$), $\lim_{m \to \infty} s_m =\frac14 \sum_{n=2}^{5} \frac1{n} $.

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  • $\begingroup$ This is where I have already gotten. I just don't know how to apply (a/1-r) to solve. $\endgroup$ Mar 1, 2015 at 1:53
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    $\begingroup$ Dropping the $1/4$, we have $(1/2 - 1/6) + (1/3 - 1/7) + (1/4 - 1/8) + (1/5 - 1/9) + (1/6 - 1/10) + (1/7 - 1/11) + (1/8 - 1/12) + ...$ Note that starting with $1/6$, terms cancel. So all you are left with is the first four terms: $1/2 + 1/3 + 1/4 + 1/5$ $\endgroup$
    – Simon S
    Mar 1, 2015 at 1:55
  • $\begingroup$ I added a complete explanation. $\endgroup$ Mar 1, 2015 at 2:09
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you can write $$\frac{4}{(n+1)(n+5)} = \frac{1}{n+1} - \frac{1}{n+5}$$ writing $n = 0, 1, 2, \cdots$ we get

$\begin{align} \frac4{1\cdot 5} &= \frac{1}{1} - \frac{1}{5}\\ \frac4{2\cdot 6} &= \frac{1}{2} - \frac{1}{6}\\ \frac4{3 \cdot 7} &= \frac{1}{3} - \frac{1}{7}\\ \frac4{4\cdot 8} &= \frac{1}{4} - \frac{1}{8}\\ \frac4{5 \cdot 9} &= \frac{1}{5} - \frac{1}{9}\\ \cdots \end{align}$

$$\sum_{n = 0}^{n = \infty}\frac{4}{(n+1)(n+5)}=1+\frac12 +\frac13+\frac14 $$

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