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Calculate the sum of the series $$\sum_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$$

I have tried partial fraction decomposition.

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

Is this correct? What is the sum?

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    $\begingroup$ are you sure the sum is over $n$, not another index? $\endgroup$ – Alex Dec 7 '16 at 23:25
  • $\begingroup$ no, im sorry is infinite $\endgroup$ – Diego Gomez Dec 7 '16 at 23:26
  • $\begingroup$ The denominator in the first sum should be $n+1$ multiplied by a constant. I think the denominator in the rightmost term reads $2(n+4)^2$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 7 '16 at 23:28
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    $\begingroup$ Note that for every $N$, $$\sum_{n=1}^N\frac{1}{4(n+2)}- \sum_{n=1}^N\frac{1}{4(n+4)}-\sum_{n=1}^N\frac{1}{2(n+4)^2}$$ is $$\frac1{4\cdot3}+\frac1{4\cdot4}-\frac1{4(N+3)}-\frac1{4(N+4)}-\frac12\sum_{n=1}^{N+4}\frac{1}{n^2}+\frac1{2\cdot1^2}+\frac1{2\cdot2^2}+\frac1{2\cdot3^2}+\frac1{2\cdot4^2}$$ hence the desired limit is $$\frac1{4\cdot3}+\frac1{4\cdot4}-\frac12\sum_{n=1}^{\infty}\frac{1}{n^2}+\frac1{2\cdot1^2}+\frac1{2\cdot2^2}+\frac1{2\cdot3^2}+\frac1{2\cdot4^2}$$ Can you finish? $\endgroup$ – Did Dec 7 '16 at 23:44
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    $\begingroup$ @Did Should become the answer. $\endgroup$ – Simply Beautiful Art Dec 7 '16 at 23:44
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Your partial fraction decomposition looks OK, but it should be written as

$$\sum_{n=1}^\infty\left({1\over4(n+2)}-{1\over4(n+4)}\right)-\sum_{n=1}^\infty{1\over2(n+4)^2}$$

instead of being split into three infinite series. That's because $\sum_{n=1}^\infty{1\over4(n+2)}$ and $\sum_{n=1}^\infty{1\over4(n+4)}$ are each divergent. But combined they give the convergent, telescoping series

$$\left({1\over4\cdot3}-{1\over4\cdot5}\right)+\left({1\over4\cdot4}-{1\over4\cdot6}\right)+\left({1\over4\cdot5}-{1\over4\cdot7}\right)+\left({1\over4\cdot6}-{1\over4\cdot8}\right)+\cdots\\={1\over4\cdot3}+{1\over4\cdot4}={7\over48}$$

The other series you need to recognize as

$${1\over2}\left({1\over5^2}+{1\over6^2}+\cdots\right)={1\over2}\sum_{n=1}^\infty{1\over n^2}-{1\over2}\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}\right)={1\over2}\left(\pi^2\over6\right)-{1\over2}\left(205\over144\right)$$

The trick is understanding the hat that the $\pi^2/6$ rabbit came from, but I'm assuming you've seen it somewhere.

I'll leave it to you to put the pieces together.

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First note that your series has the same convergence as $\sum \frac{1}{n^3}$ by the limit comparison test. And the latter series converges absolutely by the $p$-series test or the integral test, so therefore so does your series.

Next, to find the value the sum converges to, trying partial fractions, we get

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

If we break it into three series as you have attempted, then all three are divergent, and difference of divergent is indeterminate, so we cannot proceed. Instead, treat the first two terms together, and note that $\sum_{n=1}^\infty \frac{1}{n+2}-\frac{1}{n+4}$ is a telescoping series, its sum will converge to $1/3+1/4$, the uncanceled parts of the first two terms. For the final term, rememeber by the Basel problem $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$ so by reindexing we have $\sum_{n=5}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{1}{(n+4)^2}=\frac{\pi^2}{6}-\frac{1}{16}-\frac{1}{9}-\frac{1}{4}-1$.

Putting it all together we have

$$\sum_{n=1}^\infty\frac{1}{(n+2)(n+4)(n+4)}=\frac{1}{4}\left(\sum_{n=1}^\infty\frac{1}{n+2}-\frac{1}{n+4}\right)-\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(n+4)^2} \\=\frac{1}{4}\left(\frac{1}{3}+\frac{1}{4}\right)-\frac{1}{2}\left(\frac{\pi^2}{6}-\frac{1}{16}-\frac{1}{9}-\frac{1}{4}-1\right) = \frac{7}{48} -\frac{\pi^2}{12}+\frac{205}{288}=\frac{247}{288}-\frac{\pi^2}{12} $$

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  • $\begingroup$ @SimpleArt Although OP didn't write it, I assumed the question was about infinite series. Otherwise the question is kind of trivial, no? $\endgroup$ – ziggurism Dec 7 '16 at 23:27
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    $\begingroup$ I'm sorry, the OP has clarified that for us. :) On the other hand, the OP also asks to actually calculate the sum. $\endgroup$ – Simply Beautiful Art Dec 7 '16 at 23:28
  • $\begingroup$ @SimpleArt probably we should get further clarification from OP $\endgroup$ – ziggurism Dec 7 '16 at 23:30
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    $\begingroup$ Well the OP asks for the sum in the last sentence. $\endgroup$ – Simply Beautiful Art Dec 7 '16 at 23:34
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    $\begingroup$ The OP has an $(n+2)$ instead of an $(n+1)$. $\endgroup$ – Barry Cipra Dec 8 '16 at 0:05

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