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Problem: Find the sum of the series $\sum_{k=1}^{\infty} \frac {1}{(k)(k+2)(k+4)}$.

Thoughts

I first tried to write out the series to detect some kind of a pattern, I suspect I need a way of rewriting the expression $\frac {1}{(k)(k+2)(k+4)}$ , but not sure how to proceed. To help visualize it looks like:

$\frac {1} {(1)(3)(5)} + \frac {1} {(2)(4)(6)} + \frac {1} {(3)(5)(7)} + ...$

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$$\frac{1}{k(k+2)(k+4)} = \frac{1}{8} \left( \left( \frac{1}{k} - \frac{1}{k+2} \right) - \left( \frac{1}{k+2} - \frac{1}{k+4} \right) \right).$$

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An alternative approach: $$\begin{eqnarray*} \sum_{k\geq 1}\frac{1}{k(k+2)(k+4)}&=&\frac{1}{8}\sum_{k\geq 1}\left(\frac{1}{k}-\frac{2}{k+2}+\frac{1}{k+4}\right)\\&=&\frac{1}{8}\sum_{k\geq 1}\int_{0}^{1}(1-2x^2+x^4)x^{k-1}\,dx\\&=&\frac{1}{8}\int_{0}^{1}\frac{(1-x^2)^2}{1-x}\,dx\\&=&\frac{1}{8}\int_{0}^{1}(1+x-x^2-x^3)\,dx\\&=&\frac{1}{8}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=\color{red}{\frac{11}{96}}.\end{eqnarray*}$$

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  • $\begingroup$ Ha, you and your much less trivial solutions always give me a good laugh. $\endgroup$ – Simply Beautiful Art Feb 8 '17 at 2:10
  • $\begingroup$ Very nice solution (+1). Can this method be used for all partial fraction summations? $\endgroup$ – hypergeometric Feb 8 '17 at 8:53
  • $\begingroup$ @hypergeometric it looks like so. $\endgroup$ – Simply Beautiful Art Feb 8 '17 at 14:07
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Telescoping series plus some PFD:

$$\frac1{k(k+2)(k+4)}=\frac{1/8}{k(k+2)}-\frac{1/8}{(k+2)(k+4)}$$

Thus, we have

$$\begin{align}S&=\frac{1/8}{1\cdot3}-\color{#4488dd}{\frac{1/8}{3\cdot5}}\\&+\frac{1/8}{2\cdot4}-\color{#44aa88}{\frac{1/8}{4\cdot6}}\\&+\color{#4488dd}{\frac{1/8}{3\cdot5}}-\color{#dd8822}{\frac{1/8}{5\cdot7}}\\&+\color{#44aa88}{\frac{1/8}{4\cdot6}}-\color{#dd5555}{\frac{1/8}{6\cdot8}}\\&+\dots\end{align}$$

Notice each term on the right cancels with a term on the left two lines below it, giving

$$S=\frac{1/8}{1\cdot3}+\frac{1/8}{2\cdot4}=\frac{11}{96}$$

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