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I am trying to calculate the following series:

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

and I managed to reduce it to this term

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

And here I am stuck. I tried writing down a few partial sums but I can't see the pattern, $\frac{1}{2}-\frac{1}{2}+\frac{1}{5}+\frac{1}{4}-\frac{1}{3}+\frac{1}{8}+...$ I cant seem to find a closed formula that we can calculate for $S_n$

How should I go about solving this question

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Hint: $$\frac{1}{n(n+1)(n+2)}=\frac{1/2}{n(n+1)}-\frac{1/2}{(n+1)(n+2)}$$

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Consider

$$f(x) = \sum_{n=1}^{\infty} \frac{x^{n+2}}{n (n+1)(n+2)} $$

Then

$$f''(x) = -\log{(1-x)}$$

$$f'(x) = (1-x) \log{(1-x)} +x $$

$$f(x) = -\frac14 [x (2-x) - 2 (1-x)^2 \log{(1-x)}] + \frac12 x^2$$

The sum is then $f(1) = 1/4$.

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  • $\begingroup$ @Timbuc: thanks...didn't even need that line anyway. $\endgroup$ – Ron Gordon Oct 31 '14 at 20:00
  • $\begingroup$ True indeed, @Ron . :) $\endgroup$ – Timbuc Oct 31 '14 at 20:01
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    $\begingroup$ Excuse me, this is a very good method but I don't understand why the derivatives are correct. I think that $f'(x)= \sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$ $\endgroup$ – Oria Gruber Oct 31 '14 at 20:12
  • $\begingroup$ @OriaGruber: that's right. The sum works out nicely that way. Then realize that $$\sum_{n=1}^{\infty} \frac{x^n}{n} = -\log{(1-x)}$$ $\endgroup$ – Ron Gordon Oct 31 '14 at 20:14
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    $\begingroup$ ahhh I see...thats the taylor expansion...very nice. clever. thank you Ron. $\endgroup$ – Oria Gruber Oct 31 '14 at 20:24
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$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}=\\\frac{1}{2n}-\frac{2}{2n+2}+\frac{1}{2n+4}=\\(\frac{1}{2n}-\frac{1}{2n+2})+(-\frac{1}{2n+2}+\frac{1}{2n+4})=\\(\frac{1}{2n}-\frac{1}{2n+2})-(\frac{1}{2n+2}-\frac{1}{2n+4})=\\\frac{1}{2}((\frac{1}{n}-\frac{1}{n+1})-(\frac{1}{n+1}-\frac{1}{n+2}))=\\ $$

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Your partial sums are the same as those of $\frac 1 2 H_n-H_{n+1}+\frac 1 2 H_{n+2}$ minus some terms. Find those terms and use those harmonic sums tend to 0 in the limit. The answer is $1/4$. More generally, it happens that $$\binom{n+k-1}{k}^{-1}=\frac{k!}{n(n+1)\cdots (n+k)}=\sum_{j=0}^k \binom kj \frac{(-1)^j}{n+j}$$

This means that $$\sum_{n=1}^N\binom{n+k-1}k^{-1}=\sum_{j=0}^k\binom kj(-1)^j (H_{N+j}-H_j)$$

Using this and the fact that $\sum_{j=0}^k\binom kj(-1)^{j+1} H_j=\frac 1 k$ one gets $$\sum_{n\geqslant 1}\binom {n+k-1}k^{-1}=\frac 1 k$$

or $$\sum_{n\geqslant 1}\frac {1}{n(n+1)\cdots (n+k)}=\frac 1k\frac 1 {k!}$$

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