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Can you guys give me a hint on evaluating $$\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)}?$$ I have tried partial fractions but the series is not telescopic (at least I cannot see it)...

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Hint: In addition to the answers, try the Comparison test. –  Amzoti Dec 18 '12 at 1:21
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3 Answers

up vote 10 down vote accepted

Hint:

$$\frac{1}{n(n+2)(n+4)}=\frac{1}{4}\frac{n+4-n}{n(n+2)(n+4)}=\frac{1}{4}[\frac{1}{n(n+2)}-\frac{1}{(n+2)(n+4)}]$$

In general:

$$\frac{1}{n(n+d)...(n+kd)}=\frac{1}{kd}\left[\frac{1}{n(n+d)...(n+(k-1)d)}-\frac{1}{(n+d)(n+2d)...(n+kd)}\right]$$

Now if you let $a_n=\frac{-1}{kd}(\frac{1}{n(n+d)...(n+(k-1)d)})$, we find that: $$\frac{1}{n(n+d)...(n+kd)}=a_{n+d}-a_n$$ Thus, the sum $\sum_{n=1}^{\infty}\frac{1}{n(n+d)...(n+kd)}$ is a telescoping sum.

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(+1) Nice answer. –  Mhenni Benghorbal Jan 2 '13 at 2:43
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HINT: $\dfrac{1}{n(n+2)(n+4)} = \dfrac{1}{8 n}-\dfrac{1}{4(n+2)}+\dfrac{1}{8 (n+4)}.$

Now write the terms as integrals via $ \int^1_0 x^k dx = \dfrac{1}{k+1}$ and interchange integral and summation. You will have a geometric series inside which you can evaluate, and then you can evaluate the remaining integral.

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(+1) Nice answer. –  Mhenni Benghorbal Jan 2 '13 at 2:44
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$$\dfrac{1}{n(n+2)(n+4)} = \dfrac{1}{8 n}-\dfrac{1}{4(n+2)}+\dfrac{1}{8 (n+4)}= \left( \dfrac{1}{8 n}-\dfrac{1}{8(n+2)} \right)- \left(\dfrac{1}{8(n+2)}-\dfrac{1}{8 (n+4)} \right)$$

and each bracket leads to a telescopic sum...

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