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How can I evaluate this?

$$\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)} = \frac{1}{1\cdot3\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{3\cdot5\cdot7}+ \frac{1}{4\cdot6\cdot8}+\cdots$$

I have tried:

$$\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{2\cdot4\cdot6}+ \frac{1}{4\cdot6\cdot8}+\cdots = \frac{1}{3\cdot5}\left(1+\frac{1}{7}\right)+\frac{1}{4\cdot6}\left(\frac{1}{2}+\frac{1}{8}\right)+\cdots$$

and so on... Been stuck for a while. Result should be $\dfrac{11}{96}$

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  • $\begingroup$ $\frac1{1\cdot3\cdot5}=\frac14(\frac1{1\cdot3}-\frac1{3\cdot5})$ $\endgroup$
    – Empy2
    Commented Sep 13, 2015 at 15:30

5 Answers 5

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HINT you need to do a partial fraction decomposition. When you start writing out the terms you realise you have terms cancelling (it's a telescoping series). I would write it out for you but it's more fun for you to see it for yourself. :)

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With *partial fractions:$$\frac1{k(k+2)(k+4)}=\frac18\biggl(\frac1k-\frac2{k+2}+\frac1{k+4}\biggr),$$ we get a telescoping sum which simplifies to: $$\frac18\biggl(1+\frac12-\frac13-\frac14-\frac1{n+1}-\frac1{n+2}+\frac1{n+3}+\frac1{n+4}\biggr).$$

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This might be overkill but,\begin{align} \sum_{n=1}^{\infty}\frac{1}{n(n+2)(n+4)} &=\frac18\sum_{n=1}^{\infty}\left(\frac{1}{n+4}+\frac{1}{ n} -\frac{2}{n+2}\right)\tag{1}\\ &=\frac18\sum_{n=1}^{\infty}\int_{0}^{1}\left(x^{n+3}+x^{n-1}-2x^{n+1}\right)\,\mathrm dx\tag{2}\\ &=\frac18\int_{0}^{1}\sum_{n=1}^{\infty}\left(x^{n+3}+x^{n-1}-2x^{n+1}\right)\,\mathrm dx\tag{3}\\ &=\frac{1}{8}\int_{0}^{1}\left(\frac{x^{4}}{1-x}+\frac{1}{1-x}-\frac{2x^2}{1-x}\right)\,\mathrm dx\tag{4}\\ &=\frac18\int_{0}^{1}\frac{(x^2-1)^2}{1-x}\,\mathrm dx\tag{5}\\ &=\frac18\int_{0}^{1}\left(1+x-x^2-x^3\right) \,\mathrm dx\tag{6}\\ &=\frac18\Bigg[x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}\Bigg]_{0}^{1}\tag{7}\\ &=\frac{11}{96}\tag{8}\\ \end{align}

$$\sum_{n=1}^{\infty}\frac{1}{n(n+2)(n+4)}=\frac{11}{96}$$

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  • $\begingroup$ Could you explain a bit why you can do those manipulations with the integrals? I'd love to understand this :-). $\endgroup$ Commented Sep 13, 2015 at 15:32
  • $\begingroup$ @YoTengoUnLCD I'll try to find a webpage explaining that. $\endgroup$ Commented Sep 13, 2015 at 15:37
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HINT:

$$\displaystyle\dfrac4{n(n+2)(n+4)}=\dfrac{n+4-n}{n(n+2)(n+4)}=\dfrac1{n(n+2)}-\dfrac1{(n+2)(n+4)}$$ $$=g(n)-g(n+2)$$

where $g(m)=\dfrac1{m(m+2)}$

See Telescoping series

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  • $\begingroup$ The second decomposition (the incomplete one) is not needed. $\endgroup$
    – Did
    Commented Sep 13, 2015 at 14:43
  • $\begingroup$ @Did, Agreed. Thanks for your observation. $\endgroup$ Commented Sep 13, 2015 at 14:45
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$$ \frac{1}{n(n+2)(n+4)}=\frac{1}{8}(\frac{1}{n}-\frac{1}{n+2})-\frac{1}{8}(\frac{1}{(n+2}-\frac{1}{n+4})$$

we know $$\sum_{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\sum_{n=1}^{\infty }\frac{1}{n+2}$$ $$\sum_{n=1}^{\infty }\frac{1}{n+2}=\frac{1}{3}+\frac{1}{4}+\sum_{n=1}^{\infty }\frac{1}{n+4}$$ $$\frac{1}{8}(\frac{1}{1}+\frac{1}{2})-\frac{1}{8}(\frac{1}{3}+\frac{1}{4})=\frac{11}{96}$$

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